MT2176 Further Calculus - Limits of Functions Quiz

Question 1: What is the limit of the function $f(x) = \frac{x^2 - 9}{x - 3}$ as $x$ approaches 3?

A. 0 B. 3 C. 6 D. The limit does not exist.

Explanation: We can factor the numerator as $(x - 3)(x + 3)$. For $x \neq 3$, we can simplify the function to $f(x) = x + 3$. As $x$ approaches 3, $f(x)$ approaches $3 + 3 = 6$. This illustrates finding a limit by simplifying the expression. (Source: Wrede, 'Advanced Calculus', Chapter 3; MT2176 Subject Guide, Topic: Limits of functions).

Question 2: A function $f(x)$ is defined as $f(x) = \frac{\sin(x)}{x}$ for $x \neq 0$ and $f(0) = 1$. Is this function continuous at $x = 0$?

A. Yes B. No, because the limit does not exist. C. No, because f(0) is not defined. D. No, because the limit is not equal to f(0).

Explanation: A function is continuous at a point $c$ if $\lim_{x \to c} f(x) = f(c)$. We know from calculus that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$. Since $f(0)$ is defined as 1, the limit equals the function's value at the point, making it continuous. (Source: Wrede, 'Advanced Calculus', Chapter 3; MT2176 Subject Guide, Topic: Continuity).

Question 3: Using the first principles (limit definition) of the derivative, what is the derivative of $f(x) = 3x^2 + 2$?

A. $6x$ B. $3x$ C. $6x + 2$ D. $3x^2$

Explanation: The derivative is defined as $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$. Substituting $f(x) = 3x^2 + 2$, we get $\lim_{h \to 0} \frac{3(x+h)^2 + 2 - (3x^2 + 2)}{h} = \lim_{h \to 0} \frac{3(x^2 + 2xh + h^2) - 3x^2}{h} = \lim_{h \to 0} \frac{6xh + 3h^2}{h} = \lim_{h \to 0} (6x + 3h) = 6x$. (Source: Wrede, 'Advanced Calculus', Chapter 4; MT2176 Subject Guide, Topic: Definition of the derivative).

Question 4: Evaluate $\lim_{x \to 0} \frac{e^x - 1}{\sin(x)}$.

A. 0 B. 1 C. e D. The limit does not exist.

Explanation: As $x \to 0$, both the numerator and the denominator approach 0, resulting in the indeterminate form 0/0. Applying L'Hôpital's rule, we differentiate the numerator and the denominator: $\lim_{x \to 0} \frac{e^x}{\cos(x)} = \frac{e^0}{\cos(0)} = \frac{1}{1} = 1$. (Source: Wrede, 'Advanced Calculus', Chapter 4; MT2176 Subject Guide, Topic: L'Hôpital's rule).

Question 5: Use a Maclaurin series to evaluate the limit $\lim_{x \to 0} \frac{\cos(x) - 1 + x^2/2}{x^4}$.

A. 1/24 B. 1/12 C. 1/2 D. 0

Explanation: The Maclaurin series for $\cos(x)$ is $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$. Substituting this into the expression gives: $\lim_{x \to 0} \frac{(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots) - 1 + \frac{x^2}{2}}{x^4} = \lim_{x \to 0} \frac{\frac{x^4}{4!} - \dots}{x^4} = \frac{1}{4!} = \frac{1}{24}$. (Source: Wrede, 'Advanced Calculus', Chapter 11; MT2176 Subject Guide, Topic: Taylor series).

Question 6: What is the relationship between continuity and differentiability for a function at a point?

A. If a function is continuous, it must be differentiable. B. If a function is differentiable, it must be continuous. C. A function is continuous if and only if it is differentiable. D. There is no relationship between the two concepts.

Explanation: Differentiability is a stronger condition than continuity. A function can be continuous at a point without being differentiable (e.g., $f(x) = |x|$ at $x=0$), but if it is differentiable at a point, it must be continuous there. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 7: Find the limit: $\lim_{x \to \infty} \frac{3x^2 - 2x + 5}{7x^2 + 4x - 1}$.

A. 3/7 B. 5/-1 C. 0 D. $\infty$

Explanation: To find the limit at infinity for a rational function, we can divide the numerator and the denominator by the highest power of $x$, which is $x^2$. This gives $\lim_{x \to \infty} \frac{3 - 2/x + 5/x^2}{7 + 4/x - 1/x^2} = \frac{3-0+0}{7+0-0} = \frac{3}{7}$. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 8: The function $f(x) = \frac{1}{x-5}$ has a discontinuity at $x=5$. What type of discontinuity is it?

A. Removable discontinuity B. Jump discontinuity C. Infinite discontinuity D. Oscillating discontinuity

Explanation: As $x$ approaches 5, the denominator approaches 0, causing the function value to approach $\infty$ or $-\infty$. This is characteristic of an infinite discontinuity. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 9: What does Taylor's theorem provide that a simple Taylor series expansion does not?

A. A way to find the derivative of a function. B. A formula for the exact value of the function. C. An expression for the remainder term, which can be used to bound the error of the approximation. D. A method to guarantee convergence of the series.

Explanation: Taylor's theorem provides the polynomial approximation plus a remainder term (like the Lagrange or Cauchy form), which allows for the analysis of the error or the difference between the function and its polynomial approximation. (Source: MT2176 Subject Guide, Topic: Taylor's theorem).

Question 10: Evaluate $\lim_{x \to 0} \frac{\tan(x) - x}{x^3}$.

A. 1/3 B. 1/2 C. 1 D. 0

Explanation: This is an indeterminate form 0/0. Applying L'Hôpital's rule: $\lim_{x \to 0} \frac{\sec^2(x) - 1}{3x^2}$. This is still 0/0. Applying the rule again: $\lim_{x \to 0} \frac{2\sec(x)(\sec(x)\tan(x))}{6x} = \lim_{x \to 0} \frac{\sec^2(x)\tan(x)}{3x}$. Using the known limit $\lim_{u \to 0} \frac{\tan u}{u} = 1$, this becomes $\frac{1}{3} \lim_{x \to 0} \sec^2(x) \cdot 1 = \frac{1}{3}$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 11: If $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$, what is $\lim_{x \to c} [f(x)g(x)]$?

A. L + M B. L - M C. LM D. L/M

Explanation: This is the limit of a product rule. The limit of the product of two functions is the product of their limits, provided the limits exist. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 12: The function $f(x) = \begin{cases} x^2 & x < 1 \ 3-x & x \ge 1 \end{cases}$ is:

A. Continuous at x=1 B. Differentiable at x=1 C. Discontinuous at x=1 D. Undefined at x=1

Explanation: We check the left-hand and right-hand limits at x=1. $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1$. $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3-x) = 2$. Since the left-hand and right-hand limits are not equal, the overall limit does not exist, and the function is discontinuous at x=1. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 13: What is the derivative of $f(x) = \ln(x)$ from first principles?

A. $x$ B. $1/x$ C. $e^x$ D. $\ln(x)$

Explanation: The derivative is $\lim_{h \to 0} \frac{\ln(x+h) - \ln(x)}{h} = \lim_{h \to 0} \frac{1}{h} \ln(\frac{x+h}{x}) = \lim_{h \to 0} \ln((1 + \frac{h}{x})^{1/h})$. Let $u = h/x$. As $h \to 0$, $u \to 0$. The expression becomes $\frac{1}{x} \ln(\lim_{u \to 0}(1+u)^{1/u}) = \frac{1}{x} \ln(e) = \frac{1}{x}$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 14: Evaluate $\lim_{x \to \infty} (1 + \frac{a}{x})^{bx}$.

A. $e^{ab}$ B. $e^{a/b}$ C. $e^{b/a}$ D. $e^{a+b}$

Explanation: Let $y = (1 + \frac{a}{x})^{bx}$. Then $\ln(y) = bx \ln(1 + \frac{a}{x})$. We evaluate $\lim_{x \to \infty} \frac{\ln(1+a/x)}{1/(bx)}$. This is a 0/0 indeterminate form. Using L'Hôpital's rule, we get $\lim_{x \to \infty} \frac{\frac{1}{1+a/x} \cdot (-a/x^2)}{-1/(bx^2)} = \lim_{x \to \infty} \frac{ab}{1+a/x} = ab$. Since $\lim \ln(y) = ab$, we have $\lim y = e^{ab}$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 15: The first-order Taylor polynomial for $f(x) = \sqrt{x}$ around $x=4$ is:

A. $2 + \frac{1}{4}(x-4)$ B. $2 + \frac{1}{2}(x-4)$ C. $4 + \frac{1}{4}(x-4)$ D. $2 - \frac{1}{8}(x-4)$

Explanation: The first-order Taylor polynomial is $P_1(x) = f(a) + f'(a)(x-a)$. Here $a=4$, $f(x) = \sqrt{x}$, so $f(4)=2$. The derivative is $f'(x) = \frac{1}{2\sqrt{x}}$, so $f'(4) = \frac{1}{4}$. Thus, $P_1(x) = 2 + \frac{1}{4}(x-4)$. (Source: MT2176 Subject Guide, Topic: Taylor series).

Question 16: If a function is continuous on a closed interval $[a, b]$, what can be said about its differentiability?

A. It must be differentiable on $(a, b)$. B. It must be differentiable on $[a, b]$. C. It may not be differentiable anywhere on $(a, b)$. D. It must be differentiable at $x=a$ and $x=b$.

Explanation: Continuity does not imply differentiability. A classic example is the Weierstrass function, which is continuous everywhere but differentiable nowhere. A simpler example is $f(x)=|x|$ which is continuous at $x=0$ but not differentiable there. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 17: What is $\lim_{x \to 2} \frac{x^2 - 5x + 6}{x^2 - 4}$?

A. -1/4 B. 1/4 C. 0 D. 1

Explanation: Factoring the numerator and denominator gives $\lim_{x \to 2} \frac{(x-2)(x-3)}{(x-2)(x+2)} = \lim_{x \to 2} \frac{x-3}{x+2} = \frac{2-3}{2+2} = -\frac{1}{4}$. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 18: A function is continuous on $(a,b)$ if it is continuous at every point in $(a,b)$. What additional conditions are needed for it to be continuous on the closed interval $[a,b]$?

A. It must be differentiable at a and b. B. The limits at a and b must be finite. C. The right-hand limit at a must equal f(a), and the left-hand limit at b must equal f(b). D. The function must be bounded on $[a,b]$.

Explanation: For continuity on a closed interval, the function must be continuous on the open interval, and it must be continuous from the right at the left endpoint and continuous from the left at the right endpoint. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 19: The derivative of a function at a point represents the:

A. Area under the curve at that point. B. Value of the function at that point. C. Slope of the tangent line to the curve at that point. D. Overall change of the function.

Explanation: The derivative at a point is geometrically interpreted as the slope of the line tangent to the function's graph at that point. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 20: Evaluate $\lim_{x \to 0} \frac{x^2}{1 - \cos^2(x)}$.

A. 0 B. 1/2 C. 1 D. 2

Explanation: Using the identity $\sin^2(x) = 1 - \cos^2(x)$, the limit becomes $\lim_{x \to 0} \frac{x^2}{\sin^2(x)} = \lim_{x \to 0} (\frac{x}{\sin x})^2 = (\lim_{x \to 0} \frac{x}{\sin x})^2 = 1^2 = 1$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 21: The Maclaurin series for $f(x) = (1+x)^k$ is $1 + kx + \frac{k(k-1)}{2}x^2 + \dots$. What is the limit of $\frac{(1+x)^k - 1}{x}$ as $x \to 0$?

A. 1 B. k C. k/2 D. 0

Explanation: This is the definition of the derivative of $f(x)=(1+x)^k$ at $x=0$. The derivative is $f'(x) = k(1+x)^{k-1}$, so $f'(0) = k$. Alternatively, using the series: $\lim_{x \to 0} \frac{(1+kx+\dots) - 1}{x} = \lim_{x \to 0} \frac{kx+\dots}{x} = k$. (Source: MT2176 Subject Guide, Topic: Taylor series).

Question 22: If $\lim_{x \to c} f(x) = \infty$ and $\lim_{x \to c} g(x) = \infty$, what is $\lim_{x \to c} [f(x) - g(x)]$?

A. 0 B. $\infty$ C. $-\infty$ D. Indeterminate

Explanation: The form $\infty - \infty$ is indeterminate. The limit could be anything depending on the functions $f(x)$ and $g(x)$. For example, if $f(x) = x^2$ and $g(x) = x$, the limit is $\infty$. If $f(x) = x$ and $g(x) = x^2$, the limit is $-\infty$. If $f(x) = x+5$ and $g(x) = x$, the limit is 5. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 23: Is the function $f(x) = \sin(1/x)$ continuous at $x=0$?

A. Yes, it is continuous. B. No, it has a removable discontinuity. C. No, it has an infinite discontinuity. D. No, it has an oscillating discontinuity and is not defined at x=0.

Explanation: The function is not defined at $x=0$. Furthermore, as $x$ approaches 0, $1/x$ approaches infinity, and $\sin(1/x)$ oscillates infinitely between -1 and 1. The limit does not exist, so it cannot be made continuous. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 24: The statement 'If a function is continuous on $[a, b]$, then it is differentiable on \((a, b)\' is:

A. Always true B. Always false C. Sometimes true, sometimes false D. True only if the function is a polynomial

Explanation: This statement is false in general. A counterexample is $f(x) = |x|$ on $[-1, 1]$, which is continuous but not differentiable at $x=0$. However, if the function happens to be smooth, like $f(x)=x^2$, it is both continuous and differentiable. (Source: MT2176 Subject Guide, Topic: Relationship between limits, continuity and differentiability).

Question 25: Evaluate $\lim_{x \to 0^+} x \ln(x)$.

A. 0 B. 1 C. -1 D. $-\infty$

Explanation: This is an indeterminate form $0 \cdot (-\infty)$. We can rewrite it as $\lim_{x \to 0^+} \frac{\ln(x)}{1/x}$, which is an $\infty/\infty$ form. Applying L'Hôpital's rule: $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 26: Using the Taylor expansion for $e^x$ about $x=0$, find the limit of $\frac{e^x - 1 - x}{x^2}$ as $x \to 0$.

A. 0 B. 1 C. 1/2 D. 2

Explanation: The Taylor (Maclaurin) series for $e^x$ is $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$. Substituting this gives $\lim_{x \to 0} \frac{(1 + x + x^2/2! + \dots) - 1 - x}{x^2} = \lim_{x \to 0} \frac{x^2/2 + O(x^3)}{x^2} = \frac{1}{2}$. (Source: MT2176 Subject Guide, Topic: Taylor series).

Question 27: If $f(x)$ is differentiable at a point $c$, which of the following must be true?

A. $f$ is defined at $c$. B. $\lim_{x \to c} f(x)$ exists. C. $\lim_{x \to c} f(x) = f(c)$. D. All of the above.

Explanation: Differentiability at a point implies continuity at that point. The definition of continuity requires the function to be defined at the point, the limit to exist at the point, and for these two values to be equal. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 28: What is $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$?

A. 4 B. 8 C. 12 D. 16

Explanation: This is a 0/0 indeterminate form. Using L'Hôpital's rule, we get $\lim_{x \to 2} \frac{3x^2}{1} = 3(2^2) = 12$. Alternatively, factoring the numerator as a difference of cubes gives $\frac{(x-2)(x^2-2x+4)}{x-2} = x^2-2x+4$, which approaches 12 as $x \to 2$. (Source: Wrede, 'Advanced Calculus', Chapter 3 & 4).

Question 29: A function with a 'jump' discontinuity at a point $x=c$ has which property?

A. The limit does not exist, but the function is defined at c. B. The left-hand and right-hand limits exist but are not equal. C. The limit approaches infinity. D. The function is not defined at c.

Explanation: A jump discontinuity is defined as the case where the left-hand and right-hand limits both exist as finite numbers, but they are not equal to each other. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 30: Using the definition of the derivative, what is $f'(a)$ for $f(x) = \sqrt{x}$?

A. $\frac{1}{2\sqrt{a}}$ B. $\sqrt{a}$ C. $\frac{1}{\sqrt{a}}$ D. $2\sqrt{a}$

Explanation: We evaluate $\lim_{h \to 0} \frac{\sqrt{a+h} - \sqrt{a}}{h}$. Multiplying by the conjugate gives $\lim_{h \to 0} \frac{(a+h) - a}{h(\sqrt{a+h} + \sqrt{a})} = \lim_{h \to 0} \frac{h}{h(\sqrt{a+h} + \sqrt{a})} = \frac{1}{2\sqrt{a}}$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 31: Evaluate $\lim_{x \to \infty} x \sin(1/x)$.

A. 0 B. 1 C. $\infty$ D. Does not exist.

Explanation: This is an $\infty \cdot 0$ indeterminate form. Let $u = 1/x$. As $x \to \infty$, $u \to 0^+$. The limit becomes $\lim_{u \to 0^+} \frac{\sin(u)}{u}$, which is a standard limit equal to 1. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 32: The third-order Maclaurin polynomial for $\sin(x)$ is:

A. $x - \frac{x^3}{6}$ B. $x - \frac{x^3}{3}$ C. $x + \frac{x^3}{6}$ D. $x - x^2 + x^3$

Explanation: The Maclaurin series for $\sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$. The third-order polynomial is $x - \frac{x^3}{6}$. (Source: MT2176 Subject Guide, Topic: Taylor series).

Question 33: If $\lim_{x \to c} f(x) = 5$ and the function has a removable discontinuity at $x=c$, what can we say about $f(c)$?

A. $f(c) = 5$ B. $f(c) \neq 5$ or $f(c)$ is undefined. C. $f(c)$ must be 0. D. $f(c)$ must be $\infty$.

Explanation: A removable discontinuity means the limit exists, but it is not equal to the function's value at that point, either because the function has a different value or is not defined there at all. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 34: If $f(x) = |x-2|$, is the function differentiable at $x=2$?

A. Yes, the derivative is 0. B. Yes, the derivative is 1. C. No, because the function is not continuous at x=2. D. No, because the left-hand and right-hand derivatives are not equal.

Explanation: The function is continuous at $x=2$. However, the derivative from the left is -1, and the derivative from the right is 1. Since they are not equal, the function is not differentiable at $x=2$. This is a characteristic 'sharp corner'. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 35: Evaluate $\lim_{x \to 0} (\csc x - \cot x)$.

A. 0 B. 1 C. 1/2 D. Does not exist.

Explanation: This is an $\infty - \infty$ form. Rewrite as $\lim_{x \to 0} (\frac{1}{\sin x} - \frac{\cos x}{\sin x}) = \lim_{x \to 0} \frac{1 - \cos x}{\sin x}$. This is a 0/0 form. Applying L'Hôpital's rule gives $\lim_{x \to 0} \frac{\sin x}{\cos x} = \frac{0}{1} = 0$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 36: According to Taylor's theorem, the error in approximating $f(x)$ with its nth-degree Taylor polynomial $P_n(x)$ depends on which derivative?

A. The nth derivative, $f^{(n)}(x)$ B. The (n-1)th derivative, $f^{(n-1)}(x)$ C. The (n+1)th derivative, $f^{(n+1)}(x)$ D. The first derivative, $f'(x)$

Explanation: The Lagrange form of the remainder term for the nth-degree Taylor polynomial involves the (n+1)th derivative evaluated at some point between the center of the expansion and x. (Source: MT2176 Subject Guide, Topic: Taylor's theorem).

Question 37: What is $\lim_{x \to 2} \frac{x^2 - 5x + 6}{x^2 - 4}$?

A. -1/4 B. 1/4 C. 0 D. 1

Explanation: Factoring the numerator and denominator gives $\lim_{x \to 2} \frac{(x-2)(x-3)}{(x-2)(x+2)} = \lim_{x \to 2} \frac{x-3}{x+2} = \frac{2-3}{2+2} = -\frac{1}{4}$. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 38: A function is continuous on $(a,b)$ if it is continuous at every point in $(a,b)$. What additional conditions are needed for it to be continuous on the closed interval $[a,b]$?

A. It must be differentiable at a and b. B. The limits at a and b must be finite. C. The right-hand limit at a must equal f(a), and the left-hand limit at b must equal f(b). D. The function must be bounded on $[a,b]$.

Explanation: For continuity on a closed interval, the function must be continuous on the open interval, and it must be continuous from the right at the left endpoint and continuous from the left at the right endpoint. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 39: The derivative of a function at a point represents the:

A. Area under the curve at that point. B. Value of the function at that point. C. Slope of the tangent line to the curve at that point. D. Overall change of the function.

Explanation: The derivative at a point is geometrically interpreted as the slope of the line tangent to the function's graph at that point. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 40: Evaluate $\lim_{x \to 0} \frac{x^2}{1 - \cos^2(x)}$.

A. 0 B. 1/2 C. 1 D. 2

Explanation: Using the identity $\sin^2(x) = 1 - \cos^2(x)$, the limit becomes $\lim_{x \to 0} \frac{x^2}{\sin^2(x)} = \lim_{x \to 0} (\frac{x}{\sin x})^2 = (\lim_{x \to 0} \frac{x}{\sin x})^2 = 1^2 = 1$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 41: The Maclaurin series for $f(x) = (1+x)^k$ is $1 + kx + \frac{k(k-1)}{2}x^2 + \dots$. What is the limit of $\frac{(1+x)^k - 1}{x}$ as $x \to 0$?

A. 1 B. k C. k/2 D. 0

Explanation: This is the definition of the derivative of $f(x)=(1+x)^k$ at $x=0$. The derivative is $f'(x) = k(1+x)^{k-1}$, so $f'(0) = k$. Alternatively, using the series: $\lim_{x \to 0} \frac{(1+kx+\dots) - 1}{x} = \lim_{x \to 0} \frac{kx+\dots}{x} = k$. (Source: MT2176 Subject Guide, Topic: Taylor series).

Question 42: If $\lim_{x \to c} f(x) = \infty$ and $\lim_{x \to c} g(x) = \infty$, what is $\lim_{x \to c} [f(x) - g(x)]$?

A. 0 B. $\infty$ C. $-\infty$ D. Indeterminate

Explanation: The form $\infty - \infty$ is indeterminate. The limit could be anything depending on the functions $f(x)$ and $g(x)$. For example, if $f(x) = x^2$ and $g(x) = x$, the limit is $\infty$. If $f(x) = x$ and $g(x) = x^2$, the limit is $-\infty$. If $f(x) = x+5$ and $g(x) = x$, the limit is 5. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 43: Is the function $f(x) = \sin(1/x)$ continuous at $x=0$?

A. Yes, it is continuous. B. No, it has a removable discontinuity. C. No, it has an infinite discontinuity. D. No, it has an oscillating discontinuity and is not defined at x=0.

Explanation: The function is not defined at $x=0$. Furthermore, as $x$ approaches 0, $1/x$ approaches infinity, and $\sin(1/x)$ oscillates infinitely between -1 and 1. The limit does not exist, so it cannot be made continuous. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 44: The statement 'If a function is continuous on $[a, b]$, then it is differentiable on \((a, b)\' is:

A. Always true B. Always false C. Sometimes true, sometimes false D. True only if the function is a polynomial

Explanation: This statement is false in general. A counterexample is $f(x) = |x|$ on $[-1, 1]$, which is continuous but not differentiable at $x=0$. However, if the function happens to be smooth, like $f(x)=x^2$, it is both continuous and differentiable. (Source: MT2176 Subject Guide, Topic: Relationship between limits, continuity and differentiability).

Question 45: Evaluate $\lim_{x \to 0^+} x \ln(x)$.

A. 0 B. 1 C. -1 D. $-\infty$

Explanation: This is an indeterminate form $0 \cdot (-\infty)$. We can rewrite it as $\lim_{x \to 0^+} \frac{\ln(x)}{1/x}$, which is an $\infty/\infty$ form. Applying L'Hôpital's rule: $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 46: Using the Taylor expansion for $e^x$ about $x=0$, find the limit of $\frac{e^x - 1 - x}{x^2}$ as $x \to 0$.

A. 0 B. 1 C. 1/2 D. 2

Explanation: The Taylor (Maclaurin) series for $e^x$ is $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$. Substituting this gives $\lim_{x \to 0} \frac{(1 + x + x^2/2! + \dots) - 1 - x}{x^2} = \lim_{x \to 0} \frac{x^2/2 + O(x^3)}{x^2} = \frac{1}{2}$. (Source: MT2176 Subject Guide, Topic: Taylor series).

Question 47: If $f(x)$ is differentiable at a point $c$, which of the following must be true?

A. $f$ is defined at $c$. B. $\lim_{x \to c} f(x)$ exists. C. $\lim_{x \to c} f(x) = f(c)$. D. All of the above.

Explanation: Differentiability at a point implies continuity at that point. The definition of continuity requires the function to be defined at the point, the limit to exist at the point, and for these two values to be equal. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 48: What is $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$?

A. 4 B. 8 C. 12 D. 16

Explanation: This is a 0/0 indeterminate form. Using L'Hôpital's rule, we get $\lim_{x \to 2} \frac{3x^2}{1} = 3(2^2) = 12$. Alternatively, factoring the numerator as a difference of cubes gives $\frac{(x-2)(x^2-2x+4)}{x-2} = x^2-2x+4$, which approaches 12 as $x \to 2$. (Source: Wrede, 'Advanced Calculus', Chapter 3 & 4).

Question 49: A function with a 'jump' discontinuity at a point $x=c$ has which property?

A. The limit does not exist, but the function is defined at c. B. The left-hand and right-hand limits exist but are not equal. C. The limit approaches infinity. D. The function is not defined at c.

Explanation: A jump discontinuity is defined as the case where the left-hand and right-hand limits both exist as finite numbers, but they are not equal to each other. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 50: Using the definition of the derivative, what is $f'(a)$ for $f(x) = \sqrt{x}$?

A. $\frac{1}{2\sqrt{a}}$ B. $\sqrt{a}$ C. $\frac{1}{\sqrt{a}}$ D. $2\sqrt{a}$

Explanation: We evaluate $\lim_{h \to 0} \frac{\sqrt{a+h} - \sqrt{a}}{h}$. Multiplying by the conjugate gives $\lim_{h \to 0} \frac{(a+h) - a}{h(\sqrt{a+h} + \sqrt{a})} = \lim_{h \to 0} \frac{h}{h(\sqrt{a+h} + \sqrt{a})} = \frac{1}{2\sqrt{a}}$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 51: Evaluate $\lim_{x \to \infty} x \sin(1/x)$.

A. 0 B. 1 C. $\infty$ D. Does not exist.

Explanation: This is an $\infty \cdot 0$ indeterminate form. Let $u = 1/x$. As $x \to \infty$, $u \to 0^+$. The limit becomes $\lim_{u \to 0^+} \frac{\sin(u)}{u}$, which is a standard limit equal to 1. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 52: The third-order Maclaurin polynomial for $\sin(x)$ is:

A. $x - \frac{x^3}{6}$ B. $x - \frac{x^3}{3}$ C. $x + \frac{x^3}{6}$ D. $x - x^2 + x^3$

Explanation: The Maclaurin series for $\sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$. The third-order polynomial is $x - \frac{x^3}{6}$. (Source: MT2176 Subject Guide, Topic: Taylor series).

Question 53: If $\lim_{x \to c} f(x) = 5$ and the function has a removable discontinuity at $x=c$, what can we say about $f(c)$?

A. $f(c) = 5$ B. $f(c) \neq 5$ or $f(c)$ is undefined. C. $f(c)$ must be 0. D. $f(c)$ must be $\infty$.

Explanation: A removable discontinuity means the limit exists, but it is not equal to the function's value at that point, either because the function has a different value or is not defined there at all. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 54: If $f(x) = |x-2|$, is the function differentiable at $x=2$?

A. Yes, the derivative is 0. B. Yes, the derivative is 1. C. No, because the function is not continuous at x=2. D. No, because the left-hand and right-hand derivatives are not equal.

Explanation: The function is continuous at $x=2$. However, the derivative from the left is -1, and the derivative from the right is 1. Since they are not equal, the function is not differentiable at $x=2$. This is a characteristic 'sharp corner'. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 55: Evaluate $\lim_{x \to 0} (\csc x - \cot x)$.

A. 0 B. 1 C. 1/2 D. Does not exist.

Explanation: This is an $\infty - \infty$ form. Rewrite as $\lim_{x \to 0} (\frac{1}{\sin x} - \frac{\cos x}{\sin x}) = \lim_{x \to 0} \frac{1 - \cos x}{\sin x}$. This is a 0/0 form. Applying L'Hôpital's rule gives $\lim_{x \to 0} \frac{\sin x}{\cos x} = \frac{0}{1} = 0$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 56: According to Taylor's theorem, the error in approximating $f(x)$ with its nth-degree Taylor polynomial $P_n(x)$ depends on which derivative?

A. The nth derivative, $f^{(n)}(x)$ B. The (n-1)th derivative, $f^{(n-1)}(x)$ C. The (n+1)th derivative, $f^{(n+1)}(x)$ D. The first derivative, $f'(x)$

Explanation: The Lagrange form of the remainder term for the nth-degree Taylor polynomial involves the (n+1)th derivative evaluated at some point between the center of the expansion and x. (Source: MT2176 Subject Guide, Topic: Taylor's theorem).

Question 57: What is $\lim_{x \to 3} \frac{x-3}{x^2-9}$?

A. 1/6 B. 6 C. 0 D. Does not exist

Explanation: Factoring the denominator gives $\lim_{x \to 3} \frac{x-3}{(x-3)(x+3)} = \lim_{x \to 3} \frac{1}{x+3} = \frac{1}{6}$. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 58: The function $f(x) = \lfloor x \rfloor$ (the greatest integer function) is discontinuous at:

A. All real numbers B. All integers C. All rational numbers D. No points

Explanation: The greatest integer function has a jump discontinuity at every integer, as its value changes abruptly. For example, $\lim_{x \to 2^-} \lfloor x \rfloor = 1$ but $\lim_{x \to 2^+} \lfloor x \rfloor = 2$. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 59: Using the definition of the derivative, the derivative of $f(x) = \sin(x)$ is:

A. $\cos(x)$ B. $-\cos(x)$ C. $\sin(x)$ D. $-\sin(x)$

Explanation: The derivative is $\lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h} = \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} = \lim_{h \to 0} (\sin x \frac{\cos h - 1}{h} + \cos x \frac{\sin h}{h})$. Using the known limits $\lim_{h \to 0} \frac{\sin h}{h} = 1$ and $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$, the result is $\cos(x)$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 60: Evaluate $\lim_{x \to \infty} (\sqrt{x^2+x} - x)$.

A. 0 B. 1 C. 1/2 D. $\infty$

Explanation: This is an $\infty - \infty$ form. Multiply by the conjugate: $\lim_{x \to \infty} \frac{(x^2+x) - x^2}{\sqrt{x^2+x} + x} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2+x} + x}$. Divide numerator and denominator by $x$: $\lim_{x \to \infty} \frac{1}{\sqrt{1+1/x} + 1} = \frac{1}{1+1} = \frac{1}{2}$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 61: The Taylor series is a generalization of which other series?

A. Fourier Series B. Geometric Series C. Harmonic Series D. Maclaurin Series

Explanation: A Maclaurin series is a Taylor series centered at $c=0$. Therefore, the Taylor series is the more general form. (Source: MT2176 Subject Guide).

Question 62: If $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = \infty$, what is $\lim_{x \to c} [f(x)g(x)]$?

A. 0 B. $\infty$ C. 1 D. Indeterminate

Explanation: The form $0 \cdot \infty$ is indeterminate. It can be converted to a 0/0 or $\infty/\infty$ form to be evaluated, for example, by writing it as $f(x) / (1/g(x))$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 63: A function is said to be differentiable if:

A. Its limit exists everywhere. B. It is continuous everywhere. C. Its derivative exists at every point in its domain. D. It is a polynomial.

Explanation: Differentiability is defined by the existence of the derivative at each point in the function's domain. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 64: Evaluate $\lim_{x \to 0} \frac{e^x - 1 - e^{-x}}{x}$.

A. 1 B. 2 C. -1 D. -2

Explanation: This is a 0/0 indeterminate form. Applying L'Hôpital's rule: $\lim_{x \to 0} \frac{e^x - (-e^{-x})}{1} = \frac{e^0 + e^0}{1} = 1+1=2$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 65: Using the Taylor expansion of $\ln(1-x)$ about $x=0$, find the limit of $\frac{x + \ln(1-x)}{x^2}$ as $x \to 0$.

A. 1/2 B. -1/2 C. 1 D. -1

Explanation: The Taylor series for $\ln(1-x)$ is $-x - \frac{x^2}{2} - \frac{x^3}{3} - \dots$. Substituting this gives $\lim_{x \to 0} \frac{x + (-x - x^2/2 - \dots)}{x^2} = \lim_{x \to 0} \frac{-x^2/2 + O(x^3)}{x^2} = -\frac{1}{2}$. (Source: MT2176 Subject Guide, Topic: Taylor series).

Question 66: If the derivative of a function is zero over an interval, the function must be:

A. Strictly increasing B. Strictly decreasing C. Constant D. Zero

Explanation: A derivative of zero implies that the rate of change is zero, meaning the function's value does not change over that interval. This is a consequence of the Mean Value Theorem. (Source: Ostaszewski, 'Advanced mathematical methods', Chapter 17).

Question 67: What is $\lim_{x \to -1} \frac{x^2 - 1}{x+1}$?

A. 2 B. -2 C. 0 D. Does not exist

Explanation: Factoring the numerator gives $\lim_{x \to -1} \frac{(x-1)(x+1)}{x+1} = \lim_{x \to -1} (x-1) = -1-1 = -2$. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 68: The function $f(x) = \frac{x^2-x-6}{x-3}$ has a removable discontinuity at $x=3$. What value should be assigned to $f(3)$ to make it continuous?

A. 3 B. 5 C. 0 D. 6

Explanation: We find the limit as $x \to 3$. Factoring the numerator gives $\lim_{x \to 3} \frac{(x-3)(x+2)}{x-3} = \lim_{x \to 3} (x+2) = 5$. To make the function continuous, we define $f(3)$ to be equal to this limit. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 69: The derivative of $f(x) = e^{2x}$ from first principles is:

A. $e^{2x}$ B. $2e^{2x}$ C. $e^x$ D. $2e^x$

Explanation: We evaluate $\lim_{h \to 0} \frac{e^{2(x+h)} - e^{2x}}{h} = \lim_{h \to 0} \frac{e^{2x}e^{2h} - e^{2x}}{h} = e^{2x} \lim_{h \to 0} \frac{e^{2h}-1}{h}$. Let $u=2h$. The limit becomes $e^{2x} \lim_{u \to 0} \frac{e^u-1}{u/2} = 2e^{2x} \lim_{u \to 0} \frac{e^u-1}{u} = 2e^{2x}(1) = 2e^{2x}$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 70: Evaluate $\lim_{x \to 0} (\frac{1}{x^2})^{x}$.

A. 0 B. 1 C. $\infty$ D. e

Explanation: This is an $\infty^0$ indeterminate form. Let $y = (1/x^2)^x$. Then $\ln(y) = x \ln(1/x^2) = -2x \ln(x)$. We know $\lim_{x \to 0} x \ln(x) = 0$, so $\lim_{x \to 0} \ln(y) = 0$. Therefore, $\lim_{x \to 0} y = e^0 = 1$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 71: The Taylor expansion for $f(x)$ around $x=a$ is a power series in terms of:

A. $x$ B. $a$ C. $(x-a)$ D. $(x+a)$

Explanation: A Taylor series for a function $f(x)$ about a point $x=a$ is an infinite sum of terms expressed in powers of $(x-a)$. (Source: MT2176 Subject Guide).

Question 72: If $\lim_{x \to c} f(x) = L$ and $L \neq 0$, what is $\lim_{x \to c} \frac{1}{f(x)}$?

A. $1/L$ B. $L$ C. 0 D. Does not exist

Explanation: This is an application of the limit of a quotient rule, where the numerator is the constant function 1. The limit of the reciprocal is the reciprocal of the limit, provided the limit is not zero. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 73: The function $f(x) = \frac{1}{x^2+1}$ is:

A. Continuous for all real x B. Discontinuous at x=1 C. Discontinuous at x=-1 D. Discontinuous at x=0

Explanation: The function is a rational function. The denominator $x^2+1$ is never zero for any real $x$. Therefore, the function is continuous for all real numbers. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 74: If $f(x)$ is differentiable and its tangent line at $x=c$ is horizontal, what can be said about $f'(c)$?

A. $f'(c) > 0$ B. $f'(c) < 0$ C. $f'(c) = 0$ D. $f'(c)$ is undefined.

Explanation: A horizontal tangent line has a slope of 0. Since the derivative at a point gives the slope of the tangent line, $f'(c)$ must be 0. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 75: Evaluate $\lim_{x \to \pi/2} (\sec x - \tan x)$.

A. 0 B. 1 C. 1/2 D. $\infty$

Explanation: This is an $\infty - \infty$ form. Rewrite as $\lim_{x \to \pi/2} (\frac{1}{\cos x} - \frac{\sin x}{\cos x}) = \lim_{x \to \pi/2} \frac{1 - \sin x}{\cos x}$. This is a 0/0 form. Applying L'Hôpital's rule gives $\lim_{x \to \pi/2} \frac{-\cos x}{-\sin x} = \frac{0}{1} = 0$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 76: The remainder term in Taylor's theorem is used to:

A. Find the exact value of the function. B. Determine the interval of convergence. C. Analyze the error of a Taylor polynomial approximation. D. Calculate higher-order derivatives.

Explanation: The remainder term $R_n(x)$ gives the exact difference between the function $f(x)$ and its nth-degree Taylor polynomial $P_n(x)$. Bounding this term allows us to determine the accuracy of the approximation. (Source: MT2176 Subject Guide, Topic: Taylor's theorem).

Question 77: What is $\lim_{x \to -2} \frac{x^3+8}{x+2}$?

A. 4 B. 8 C. 12 D. -4

Explanation: This is a 0/0 indeterminate form. Using L'Hôpital's rule, we get $\lim_{x \to -2} \frac{3x^2}{1} = 3(-2)^2 = 12$. Alternatively, factoring the sum of cubes in the numerator gives $\frac{(x+2)(x^2-2x+4)}{x+2} = x^2-2x+4$, which approaches 12 as $x \to -2$. (Source: Wrede, 'Advanced Calculus', Chapter 3 & 4).

Question 78: True or False: A function can have a limit at a point where it is not defined.

A. True B. False

Explanation: True. The limit of a function as $x$ approaches $c$ depends on the values of the function near $c$, not on the value of the function at $c$ itself. For example, $f(x) = \frac{x^2-4}{x-2}$ is not defined at $x=2$, but its limit as $x \to 2$ is 4. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 79: The derivative of $f(x) = \cos(x)$ from first principles is:

A. $\sin(x)$ B. $-\sin(x)$ C. $\cos(x)$ D. $-\cos(x)$

Explanation: Using the definition $\lim_{h \to 0} \frac{\cos(x+h) - \cos(x)}{h} = \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h} = \lim_{h \to 0} (\cos x \frac{\cos h - 1}{h} - \sin x \frac{\sin h}{h})$. This evaluates to $\cos(x)(0) - \sin(x)(1) = -\sin(x)$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 80: Evaluate $\lim_{x \to 0^+} (\sin x)^x$.

A. 0 B. 1 C. e D. Does not exist

Explanation: This is a $0^0$ indeterminate form. Let $y = (\sin x)^x$, so $\ln y = x \ln(\sin x)$. This is a $0 \cdot (-\infty)$ form. Rewrite as $\frac{\ln(\sin x)}{1/x}$. Applying L'Hôpital's Rule: $\lim_{x \to 0^+} \frac{\frac{\cos x}{\sin x}}{-1/x^2} = \lim_{x \to 0^+} -x^2 \cot x = \lim_{x \to 0^+} -x \frac{x}{\tan x} = (0)(1) = 0$. So, $\lim_{x \to 0^+} y = e^0 = 1$. (Source: Ostaszewski, Section 18.8).

Question 81: The Maclaurin series is a special case of the Taylor series centered at:

A. x = 1 B. x = -1 C. x = 0 D. x = $\pi$

Explanation: By definition, the Maclaurin series of a function is its Taylor series expansion around the point $c=0$. (Source: MT2176 Subject Guide, Topic: Taylor series).

Question 82: If $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$, L'Hôpital's rule states that $\lim_{x \to a} \frac{f(x)}{g(x)}$ is equal to:

A. $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ B. $\frac{f'(a)}{g'(a)}$ C. $\lim_{x \to a} (f(x)-g(x))$ D. $\lim_{x \to a} (f'(x)g(x) - f(x)g'(x))$

Explanation: L'Hôpital's rule allows us to evaluate the limit of a quotient of functions by taking the limit of the quotient of their derivatives, provided the limit exists and the initial form is indeterminate (0/0 or $\infty/\infty$). (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 83: Which statement correctly describes the function $f(x) = \frac{1}{x}$ at $x=0$?

A. It is continuous. B. It has a removable discontinuity. C. It has a jump discontinuity. D. It has an infinite discontinuity.

Explanation: The function is not defined at $x=0$, and the limits from the left ($-\infty$) and right ($+\infty$) approach infinity. This is the definition of an infinite discontinuity. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 84: If a function is differentiable, it implies continuity. Does the converse hold?

A. Yes, always. B. No, never. C. Not necessarily. D. Only for linear functions.

Explanation: Continuity is a necessary condition for differentiability, but it is not a sufficient one. A function can be continuous at a point but fail to be differentiable there, such as $f(x)=|x|$ at $x=0$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 85: Evaluate $\lim_{x \to 0} \frac{\sin(5x)}{x}$.

A. 1 B. 5 C. 1/5 D. 0

Explanation: This is a 0/0 indeterminate form. Using L'Hôpital's rule: $\lim_{x \to 0} \frac{5\cos(5x)}{1} = 5\cos(0) = 5$. Alternatively, rewrite as $5 \cdot \frac{\sin(5x)}{5x}$ and use the standard limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 86: The second-order Taylor polynomial $P_2(x)$ for a function $f(x)$ at $x=a$ provides a:

A. Linear approximation B. Quadratic approximation C. Cubic approximation D. Constant approximation

Explanation: The degree of the Taylor polynomial indicates the degree of the polynomial used to approximate the function. A second-order polynomial is a quadratic. (Source: MT2176 Subject Guide, Topic: Taylor series).

Question 87: What is $\lim_{x \to \infty} e^{-x}$?

A. 0 B. 1 C. $\infty$ D. e

Explanation: As $x$ becomes infinitely large, $-x$ becomes infinitely negative, and $e$ raised to a large negative power approaches 0. The graph of $e^{-x}$ has a horizontal asymptote at $y=0$. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 88: If $f(x)$ is continuous at $x=c$, which of the following is NOT necessarily true?

A. $f(c)$ is defined. B. $\lim_{x \to c} f(x)$ exists. C. $f(x)$ is differentiable at $x=c$. D. $\lim_{x \to c} f(x) = f(c)$.

Explanation: Continuity is a necessary condition for differentiability, but it does not guarantee it. A function can be continuous at a point without being differentiable there. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 89: The derivative of $f(x) = x^n$ (for integer $n$) from first principles involves the expansion of:

A. $(x+h)^n$ B. $x^n + h^n$ C. $(x-h)^n$ D. $x^h$

Explanation: The definition of the derivative requires calculating $f(x+h)$, which in this case is $(x+h)^n$. The binomial theorem is used to expand this term. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 90: Evaluate $\lim_{x \to 0} \frac{\arcsin(x)}{x}$.

A. 0 B. 1 C. $\pi/2$ D. $\infty$

Explanation: This is a 0/0 indeterminate form. Applying L'Hôpital's rule: $\lim_{x \to 0} \frac{1/\sqrt{1-x^2}}{1} = \frac{1/\sqrt{1-0}} = 1$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 91: If a Taylor series for $f(x)$ converges to $f(x)$ in an interval, it means that:

A. The function is a polynomial. B. The remainder term $R_n(x)$ approaches 0 as $n \to \infty$. C. The function is defined everywhere. D. All derivatives are positive.

Explanation: The convergence of a Taylor series to the function itself is guaranteed if and only if the remainder term goes to zero as the degree of the polynomial approximation goes to infinity. (Source: MT2176 Subject Guide, Topic: Taylor's theorem).

Question 92: What is $\lim_{x \to 1} \frac{x^n - 1}{x - 1}$ for a positive integer $n$?

A. 1 B. n C. n-1 D. $\infty$

Explanation: This is the definition of the derivative of $f(x) = x^n$ at $x=1$. The derivative is $f'(x) = nx^{n-1}$, so $f'(1) = n(1)^{n-1} = n$. Alternatively, L'Hôpital's rule gives $\lim_{x \to 1} \frac{nx^{n-1}}{1} = n$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 93: A 'removable' discontinuity at $x=c$ means:

A. The limit does not exist. B. The function can be made continuous by redefining $f(c)$. C. The function approaches infinity. D. The function oscillates infinitely.

Explanation: A discontinuity is removable if the limit of the function exists at that point, but the function's value is either different or undefined. Continuity can be established by setting the function's value equal to the limit. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 94: If $f(x)$ is differentiable and $g(x) = f(x+c)$, what is $g'(x)$?

A. $f'(x)$ B. $f'(x+c)$ C. $f'(c)$ D. $f(x+c)$

Explanation: This is an application of the chain rule. Let $u = x+c$. Then $g(x) = f(u)$. The derivative is $\frac{dg}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = f'(u) \cdot 1 = f'(x+c)$. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 95: Evaluate $\lim_{x \to \infty} x(e^{1/x} - 1)$.

A. 0 B. 1 C. e D. $\infty$

Explanation: This is an $\infty \cdot 0$ indeterminate form. Let $u=1/x$. As $x \to \infty$, $u \to 0$. The limit becomes $\lim_{u \to 0} \frac{e^u - 1}{u}$, which is the definition of the derivative of $e^u$ at $u=0$, which is 1. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 96: The Taylor series provides a representation of a function as an infinite sum of terms calculated from the values of the function's derivatives at a single point. This point is called the:

A. Origin B. Center of the series C. Inflection point D. Critical point

Explanation: The point $c$ around which the Taylor series is expanded is known as the center of the expansion. (Source: MT2176 Subject Guide).

Question 97: What is $\lim_{x \to 4} \frac{\sqrt{x}-2}{x-4}$?

A. 1/4 B. 1/2 C. 4 D. Does not exist

Explanation: This is a 0/0 form. Using L'Hôpital's rule: $\lim_{x \to 4} \frac{1/(2\sqrt{x})}{1} = \frac{1/(2\sqrt{4})}{1} = \frac{1}{4}$. Alternatively, multiply by the conjugate of the numerator. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 98: If a function is continuous on $(a,b)$, does it have to be bounded on $(a,b)$?

A. Yes, always. B. No, not necessarily. C. Only if the interval is finite. D. Only if the function is monotonic.

Explanation: No. For example, the function $f(x) = 1/x$ is continuous on the open interval $(0,1)$ but it is not bounded, as it approaches infinity as $x$ approaches 0. (Source: Wrede, 'Advanced Calculus', Chapter 3).

Question 99: The existence of a derivative $f'(c)$ depends on the existence of:

A. A limit B. A Taylor series C. An integral D. A vertical asymptote

Explanation: The derivative is, by its very definition, the limit of a difference quotient. If this limit does not exist, the function is not differentiable at that point. (Source: Wrede, 'Advanced Calculus', Chapter 4).

Question 100: Use Taylor series to find the limit $lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$.

The Taylor series for $e^x$ is $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$.

$$ \lim_{x \to 0} \frac{(1 + x + \frac{x^2}{2!} + \dots) - 1 - x}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{2} + \frac{x^3}{6} + \dots}{x^2} $$

$$ = \lim_{x \to 0} (\frac{1}{2} + \frac{x}{6} + \dots) = \frac{1}{2} $$

Source: Wrede, Chapter 11