MT2176 Further Calculus - Chapter 2 Quiz: Limits

Correct Answer: C

Explanation: We can factor the numerator as a difference of squares: \(x^2 - 9 = (x - 3)(x + 3)\). Then, for \(x \neq 3\), the expression simplifies to \(x + 3\). The limit is then \(\lim_{x \to 3} (x + 3) = 3 + 3 = 6\). This is an example of finding a limit by simplifying the expression. Source: Adams and Essex (2010), Section 1.2.

Correct Answer: A

Explanation: Differentiability implies continuity. If a function has a derivative at a point, it must be continuous at that point. However, the converse is not true; a function can be continuous at a point without being differentiable there (e.g., \(f(x) = |x|\) at \(x=0\)). Source: Ostaszewski (2012), Section 2.2.2.

Correct Answer: B

Explanation: To find the limit at infinity of a rational function, we can divide the numerator and the denominator by the highest power of \(x\) in the denominator, which is \(x^2\). This gives \(\lim_{x \to \infty} \frac{3 - 2/x + 1/x^2}{5 + 4/x - 7/x^2}\). As \(x \to \infty\), the terms with \(x\) in the denominator go to 0, leaving \(3/5\). Source: Adams and Essex (2010), Section 1.3.

Correct Answer: B

Explanation: For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function's value at that point must all be equal. Here, \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+2) = 3\) and \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4-x) = 3\). Also, \(f(1) = 1+2 = 3\). Since the left-hand limit, right-hand limit, and function value are all equal to 3, the function is continuous at \(x=1\). Wait, I made a mistake in the explanation. Let me re-evaluate. The left limit is 3. The right limit is 3. The function value is 3. So it is continuous. Let me check the question and options again. Ah, the question in the prompt is different from what I wrote. Let me correct the question to match the intended discontinuity. Let's change the second part of the function to \(3-x\). Let's re-evaluate with \(f(x) = \begin{cases} x+2 & \text{if } x \le 1 \ 3-x & \text{if } x > 1 \end{cases}\). \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+2) = 3\). \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3-x) = 2\). Since the left and right limits are not equal, the function is not continuous. The correct answer is B. Let's try another one. Let's use the original function from my thoughts: \(f(x) = \begin{cases} x+2 & \text{if } x \le 1 \ 4-x & \text{if } x > 1 \end{cases}\). \(\lim_{x \to 1^-} f(x) = 3\). \(\lim_{x \to 1^+} f(x) = 3\). \(f(1) = 3\). So it is continuous. Let me create a question that is discontinuous. Let's use \(f(x) = \begin{cases} x^2+1 & \text{if } x < 2 \ 2x+1 & \text{if } x \ge 2 \end{cases}\). \(\lim_{x \to 2^-} f(x) = 2^2+1 = 5\). \(\lim_{x \to 2^+} f(x) = 2(2)+1 = 5\). \(f(2) = 5\). This is also continuous. Let's try \(f(x) = \begin{cases} x^2 & \text{if } x \neq 2 \ 5 & \text{if } x = 2 \end{cases}\). \(\lim_{x \to 2} f(x) = \lim_{x \to 2} x^2 = 4\). But \(f(2)=5\). So \(\lim_{x \to 2} f(x) \neq f(2)\). This is a removable discontinuity. Let's use this for the question.

Let's re-write the question and explanation.

4. Is the function \(f(x) = \begin{cases} x^2 & \text{if } x \neq 2 \ 5 & \text{if } x = 2 \end{cases}\) continuous at \(x=2\)?

• A) Yes
• B) No, because the limit does not exist.
• C) No, because \(f(2)\) is not defined.
• D) No, because the limit of \(f(x)\) as \(x \to 2\) is not equal to \(f(2)\).

Correct Answer: B

Explanation: The definition of the derivative is \(f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\). For \(f(x) = x^3\) at \(c=2\), we have: \(f'(2) = \lim_{h \to 0} \frac{(2+h)^3 - 2^3}{h} = \lim_{h \to 0} \frac{(8 + 12h + 6h^2 + h^3) - 8}{h} = \lim_{h \to 0} \frac{12h + 6h^2 + h^3}{h} = \lim_{h \to 0} (12 + 6h + h^2) = 12\). Source: Wrede and Spiegel (2010), Chapter 4.

Correct Answer: C

Explanation: The Maclaurin series (Taylor series about 0) for \(\sin(x)\) is \(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots\). Substituting this into the limit gives: \(\lim_{x \to 0} \frac{(x - \frac{x^3}{6} + \frac{x^5}{120} - \dots) - x}{x^3} = \lim_{x \to 0} \frac{-\frac{x^3}{6} + \frac{x^5}{120} - \dots}{x^3} = \lim_{x \to 0} (-\frac{1}{6} + \frac{x^2}{120} - \dots) = -\frac{1}{6}\). Source: Ostaszewski (2012), Section 2.2.3.

Correct Answer: B

Explanation: The limit is of the indeterminate form 0/0. Applying L'Hôpital's rule, we differentiate the numerator and the denominator: \(\lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(\sin(x))} = \lim_{x \to 0} \frac{e^x}{\cos(x)} = \frac{e^0}{\cos(0)} = \frac{1}{1} = 1\). Source: Adams and Essex (2010), Section 4.3.

Correct Answer: D

Explanation: The function \(\sin(x)\) oscillates between -1 and 1 as \(x\) increases. It does not approach any single value. Therefore, the limit does not exist. Source: Ostaszewski (2012), Section 2.1.2.

Correct Answer: C

Explanation: This is the product rule for limits. The limit of a product is the product of the limits, provided the individual limits exist. Source: Adams and Essex (2010), Section 1.2, Theorem 2.

Correct Answer: D

Explanation: The function \(f(x)=|x|\) is continuous at \(x=0\) but not differentiable. The left-hand derivative is \(\lim_{h \to 0^-} \frac{|0+h|-|0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1\), while the right-hand derivative is \(\lim_{h \to 0^+} \frac{|0+h|-|0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1\). Since the left and right derivatives are not equal, the derivative does not exist. Source: Ostaszewski (2012), Section 2.2.2.

Correct Answer: A

Explanation: We use the Squeeze Theorem. We know that \(-1 \le \sin(\frac{1}{x}) \le 1\) for all \(x \neq 0\). Multiplying by \(x^2\) (which is non-negative), we get \(-x^2 \le x^2 \sin(\frac{1}{x}) \le x^2\). Since \(\lim_{x \to 0} (-x^2) = 0\) and \(\lim_{x \to 0} x^2 = 0\), by the Squeeze Theorem, \(\lim_{x \to 0} x^2 \sin(\frac{1}{x}) = 0\). Source: Adams and Essex (2010), Section 1.2.

Correct Answer: C

Explanation: For continuity at \(x=2\), we need \(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)\). \(\lim_{x \to 2^-} kx^2 = k(2^2) = 4k\). \(\lim_{x \to 2^+} (2x+k) = 2(2)+k = 4+k\). Setting them equal: \(4k = 4+k \implies 3k = 4 \implies k = 4/3\). Source: Adams and Essex (2010), Section 1.4.

Correct Answer: A

Explanation: Let \(f(x) = \ln(1+x)\). Then \(f'(x) = \frac{1}{1+x}\) and \(f''(x) = -\frac{1}{(1+x)^2}\). The first-degree Taylor polynomial is \(P_1(x) = f(0) + f'(0)x = 0 + 1x = x\). The error term (Lagrange form) is \(R_1(x) = \frac{f''(c)}{2!}x^2\) for some \(c\) between 0 and \(x\). For \(x=0.1\), the error is \(|R_1(0.1)| = |\frac{-1/(1+c)^2}{2}(0.1)^2| = \frac{0.01}{2(1+c)^2}\) where \(0 < c < 0.1\). The maximum value of this error occurs when the denominator is smallest, which is when \(c=0\). So, \(|R_1(0.1)| \le \frac{0.01}{2(1+0)^2} = 0.005\). Source: Ostaszewski (2012), Section 2.2.3.

Correct Answer: B

Explanation: This is of the form \(\infty - \infty\). We multiply by the conjugate: \(\lim_{x \to \infty} (\sqrt{x^2+x} - x) \frac{\sqrt{x^2+x} + x}{\sqrt{x^2+x} + x} = \lim_{x \to \infty} \frac{x^2+x - x^2}{\sqrt{x^2+x} + x} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2(1+1/x)} + x} = \lim_{x \to \infty} \frac{x}{x\sqrt{1+1/x} + x} = \lim_{x \to \infty} \frac{1}{\sqrt{1+1/x} + 1} = \frac{1}{\sqrt{1+0} + 1} = \frac{1}{2}\). Source: Adams and Essex (2010), Section 1.3.

Correct Answer: B

Explanation: Using the definition of the derivative: \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h} = \lim_{h \to 0} \frac{-h}{hx(x+h)} = \lim_{h \to 0} \frac{-1}{x(x+h)} = -\frac{1}{x^2}\). Source: Wrede and Spiegel (2010), Chapter 4.

Correct Answer: D

Explanation: This is of the indeterminate form \(1^\infty\). Let \(y = (1+2x)^{1/x}\). Then \(\ln y = \frac{\ln(1+2x)}{x}\). We can use L'Hôpital's rule on \(\ln y\): \(\lim_{x \to 0} \frac{\ln(1+2x)}{x} = \lim_{x \to 0} \frac{2/(1+2x)}{1} = 2\). Since \(\lim_{x \to 0} \ln y = 2\), we have \(\lim_{x \to 0} y = e^2\). Source: Adams and Essex (2010), Section 4.3.

Correct Answer: D

Explanation: The function \(f(x) = x^{1/3}\) is continuous at \(x=0\). However, its derivative is \(f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}\), which is undefined at \(x=0\) as the tangent line is vertical. The function \(x|x|\) is differentiable at \(x=0\). Source: Adams and Essex (2010), Section 2.1.

Correct Answer: B

Explanation: This is a 0/0 indeterminate form. Using L'Hôpital's rule: \(\lim_{x \to 1} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x-1)} = \lim_{x \to 1} \frac{1/x}{1} = 1\). Alternatively, this is the definition of the derivative of \(f(x) = \ln x\) at \(x=1\). Source: Ostaszewski (2012), Section 2.2.4.

Correct Answer: B

Explanation: This is the formal \(\epsilon-\delta\) definition of a limit. The condition \(0 < |x-c|\) is important because it means \(x \neq c\). The value of \(f(c)\) itself does not affect the limit. Source: Adams and Essex (2010), Section 1.5.

Correct Answer: B

Explanation: The Taylor polynomial of order \(n\) for \(f(x)\) about \(x=a\) is \(P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k\). For \(f(x)=e^x\) and \(a=0\), we have \(f^{(k)}(x) = e^x\) for all \(k\), so \(f^{(k)}(0) = 1\). \(P_2(x) = \frac{f(0)}{0!}x^0 + \frac{f'(0)}{1!}x^1 + \frac{f''(0)}{2!}x^2 = 1 + x + \frac{x^2}{2}\). Source: Ostaszewski (2012), Section 2.2.3.

Correct Answer: A

Explanation: This is an indeterminate form of type \(0 \cdot (-\infty)\). We can rewrite it as \(\lim_{x \to 0^+} \frac{\ln x}{1/x}\), which is of the form \(-\infty/\infty\). Applying L'Hôpital's rule: \(\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0\). Source: Adams and Essex (2010), Section 4.3.

Correct Answer: C

Explanation: The Extreme Value Theorem states that if a function \(f\) is continuous on a closed, bounded interval \([a, b]\), then \(f\) must attain both an absolute maximum and an absolute minimum value on \([a, b]\). Source: Adams and Essex (2010), Section 1.4, Theorem 8.

Correct Answer: A

Explanation: \(f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\). Multiply by the conjugate: \(\lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{x+h-x}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2\sqrt{x}}\). Source: Ostaszewski (2012), Section 2.2.2, Example 2.19.

Correct Answer: A

Explanation: We can use the Squeeze Theorem. Since \(-1 \le \sin x \le 1\), for \(x > 0\), we have \(-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}\). As \(x \to \infty\), both \(-\frac{1}{x}\) and \(\frac{1}{x}\) approach 0. Therefore, by the Squeeze Theorem, \(\lim_{x \to \infty} \frac{\sin x}{x} = 0\). Source: Ostaszewski (2012), Section 2.1.1, Example 2.4.

Correct Answer: C

Explanation: A fundamental theorem in calculus states that differentiability at a point implies continuity at that point. The contrapositive of this statement is that if a function is not continuous at a point, it cannot be differentiable at that point. Source: Adams and Essex (2010), Section 2.3, Theorem 1.

Correct Answer: B

Explanation: \(f(x) = \cos(x)\), so \(f(\pi/2) = \cos(\pi/2) = 0\). \(f'(x) = -\sin(x)\), so \(f'(\pi/2) = -\sin(\pi/2) = -1\). The first-order Taylor polynomial is \(P_1(x) = f(\pi/2) + f'(\pi/2)(x - \pi/2) = 0 - 1(x - \pi/2) = -(x - \pi/2)\). Source: Wrede and Spiegel (2010), Chapter 11.

Correct Answer: C

Explanation: This is a 0/0 indeterminate form. Using L'Hôpital's rule twice: \(\lim_{x \to 0} \frac{\sin(x)}{2x}\). This is still 0/0. Applying the rule again: \(\lim_{x \to 0} \frac{\cos(x)}{2} = \frac{1}{2}\). Alternatively, using the Taylor series for \(\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots\), we get \(\lim_{x \to 0} \frac{1 - (1 - x^2/2 + \dots)}{x^2} = \lim_{x \to 0} \frac{x^2/2 - \dots}{x^2} = 1/2\). Source: Ostaszewski (2012), Section 2.2.4, Example 2.23.

Correct Answer: B

Explanation: The function is the signum function, which is -1 for \(x<0\) and 1 for \(x>0\). \(\lim_{x \to 0^-} f(x) = -1\) and \(\lim_{x \to 0^+} f(x) = 1\). Since the left and right limits exist but are not equal, this is a jump discontinuity. Source: Adams and Essex (2010), Section 1.2, Example 6.

Correct Answer: D

Explanation: \(f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{(2(1+h)^2 - 3) - (2(1)^2 - 3)}{h} = \lim_{h \to 0} \frac{2(1+2h+h^2)-3 - (-1)}{h} = \lim_{h \to 0} \frac{2+4h+2h^2-2}{h} = \lim_{h \to 0} \frac{4h+2h^2}{h} = \lim_{h \to 0} (4+2h) = 4\). Source: Wrede and Spiegel (2010), Chapter 4.

Correct Answer: D

Explanation: This is the famous limit definition of the number \(e\). It's an indeterminate form of type \(1^\infty\). Source: Adams and Essex (2010), Section 1.2, Example 2.

Correct Answer: B

Explanation: Continuity does not imply differentiability. A classic counterexample is \(f(x) = |x|\) at \(x=0\). The function is continuous at 0, but it has a sharp corner, so it is not differentiable there. Source: Ostaszewski (2012), Section 2.2.2.

Correct Answer: B

Explanation: This is a 0/0 indeterminate form. Using L'Hôpital's rule: \(\lim_{x \to 0} \frac{\sec^2(x)}{1} = \sec^2(0) = 1^2 = 1\). Alternatively, rewrite as \(\lim_{x \to 0} \frac{\sin(x)}{x \cos(x)} = (\lim_{x \to 0} \frac{\sin(x)}{x})(\lim_{x \to 0} \frac{1}{\cos(x)}) = (1)(1) = 1\). Source: Adams and Essex (2010), Section 2.5.

Correct Answer: D

Explanation: The Intermediate Value Theorem states that if a function \(f\) is continuous on a closed interval \([a, b]\), then for any value \(y\) between \(f(a)\) and \(f(b)\), there is at least one number \(c\) in \([a, b]\) such that \(f(c) = y\). Continuity on the closed interval is the key hypothesis. Source: Adams and Essex (2010), Section 1.4, Theorem 9.

Correct Answer: C

Explanation: The Maclaurin series for \(e^x\) is \(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots\). Substituting this into the limit: \(\lim_{x \to 0} \frac{(1 + x + \frac{x^2}{2} + \dots) - 1 - x}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{2} + \frac{x^3}{6} + \dots}{x^2} = \lim_{x \to 0} (\frac{1}{2} + \frac{x}{6} + \dots) = \frac{1}{2}\). Source: Ostaszewski (2012), Section 2.2.3.

Correct Answer: B

Explanation: This is an indeterminate form \(\infty^0\). Let \(y = x^{1/x}\), so \(\ln y = \frac{\ln x}{x}\). We evaluate the limit of \(\ln y\) as \(x \to \infty\). This is of the form \(\infty/\infty\), so we use L'Hôpital's rule: \(\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0\). Since \(\lim_{x \to \infty} \ln y = 0\), we have \(\lim_{x \to \infty} y = e^0 = 1\). Source: Adams and Essex (2010), Section 4.3.

Correct Answer: B

Explanation: \(f'(c) = \lim_{h \to 0} \frac{\sin(c+h) - \sin(c)}{h}\). Using the angle addition formula for sine, \(\sin(c+h) = \sin(c)\cos(h) + \cos(c)\sin(h)\). The limit becomes \(\lim_{h \to 0} \frac{\sin(c)\cos(h) + \cos(c)\sin(h) - \sin(c)}{h} = \lim_{h \to 0} [\sin(c)\frac{\cos(h)-1}{h} + \cos(c)\frac{\sin(h)}{h}]\). Using the known limits \(\lim_{h \to 0} \frac{\sin(h)}{h} = 1\) and \(\lim_{h \to 0} \frac{\cos(h)-1}{h} = 0\), we get \(\sin(c)(0) + \cos(c)(1) = \cos(c)\). Source: Adams and Essex (2010), Section 2.5.

Correct Answer: C

Explanation: As \(x\) approaches 0 from either the positive or negative side, \(x^2\) is a small positive number. The reciprocal of a small positive number is a large positive number. Therefore, the limit is \(\infty\). This is an example of an infinite limit. Source: Adams and Essex (2010), Section 1.3.

Correct Answer: C

Explanation: To remove the discontinuity, we need to define \(f(2)\) to be equal to the limit of \(f(x)\) as \(x \to 2\). \(\lim_{x \to 2} \frac{x^2-4}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4\). By defining \(f(2) = 4\), the function becomes continuous. Source: Adams and Essex (2010), Section 1.4.

Correct Answer: C

Explanation: Substitute the Taylor series into the expression: \(\lim_{x \to 0} \frac{(x - x^2/2 + x^3/3 - \dots) - x}{x^2} = \lim_{x \to 0} \frac{-x^2/2 + x^3/3 - \dots}{x^2} = \lim_{x \to 0} (-1/2 + x/3 - \dots) = -1/2\). Source: Ostaszewski (2012), Section 2.2.3.

Correct Answer: A

Explanation: This is an indeterminate form \(\infty - \infty\). We combine the terms: \(\lim_{x \to \pi/2} (\frac{\sin x}{\cos x} - \frac{1}{\cos x}) = \lim_{x \to \pi/2} \frac{\sin x - 1}{\cos x}\). This is now of the form 0/0. Using L'Hôpital's rule: \(\lim_{x \to \pi/2} \frac{\cos x}{-\sin x} = \frac{\cos(\pi/2)}{-\sin(\pi/2)} = \frac{0}{-1} = 0\). Source: Wrede and Spiegel (2010), Chapter 4.

Correct Answer: A

Explanation: This is a direct application of the Intermediate Value Theorem, sometimes called Bolzano's Theorem. If \(f(a)\) and \(f(b)\) have opposite signs, then 0 is a value between \(f(a)\) and \(f(b)\). The theorem guarantees there is a \(c\) in \((a,b)\) such that \(f(c)=0\). Source: Adams and Essex (2010), Section 1.4.

Correct Answer: C

Explanation: This is the definition of the derivative of \(f(x) = e^{2x}\) at \(x=0\). Alternatively, using L'Hôpital's rule for the 0/0 form: \(\lim_{x \to 0} \frac{2e^{2x}}{1} = 2e^0 = 2\). Source: Wrede and Spiegel (2010), Chapter 4.

Correct Answer: C

Explanation: As \(x\) approaches 3, the denominator approaches 0, while the numerator is 1. This causes the function's value to grow without bound. \(\lim_{x \to 3^+} \frac{1}{x-3} = \infty\) and \(\lim_{x \to 3^-} \frac{1}{x-3} = -\infty\). This is an infinite discontinuity. Source: Adams and Essex (2010), Section 1.4.

Correct Answer: A

Explanation: \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{(5(x+h) - 3) - (5x - 3)}{h} = \lim_{h \to 0} \frac{5x+5h-3-5x+3}{h} = \lim_{h \to 0} \frac{5h}{h} = 5\). Source: Wrede and Spiegel (2010), Chapter 4.

Correct Answer: C

Explanation: We can rewrite the limit as \(\lim_{x \to \infty} (\frac{1}{(x+1)/x})^x = \lim_{x \to \infty} \frac{1}{(1 + 1/x)^x}\). Since \(\lim_{x \to \infty} (1 + 1/x)^x = e\), the limit is \(\frac{1}{e}\). Source: Adams and Essex (2010), Section 3.3.

Correct Answer: A

Explanation: For \(f(x) = \cos(x)\), the derivatives at 0 are \(f(0)=1, f'(0)=0, f''(0)=-1, f'''(0)=0\). The third-order polynomial is \(P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 = 1 + 0x - \frac{1}{2}x^2 + 0x^3 = 1 - \frac{x^2}{2}\). Source: Adams and Essex (2010), Section 4.10.

Correct Answer: C

Explanation: This is the definition of the derivative of \(f(x) = \sqrt{1+x}\) at \(x=0\). The derivative is \(f'(x) = \frac{1}{2\sqrt{1+x}}\), so \(f'(0) = 1/2\). Alternatively, using L'Hôpital's rule for the 0/0 form: \(\lim_{x \to 0} \frac{\frac{1}{2\sqrt{1+x}}}{1} = \frac{1}{2}\). Source: Ostaszewski (2012), Section 2.2.2.

Correct Answer: D

Explanation: A removable discontinuity occurs when the limit of the function at a point exists, but the function's value at that point is either different or not defined. The discontinuity can be "removed" by redefining the function at that point to be equal to the limit. Source: Adams and Essex (2010), Section 1.4.

Correct Answer: A

Explanation: This is a 0/0 form. Using L'Hôpital's rule repeatedly: \(\lim_{x \to 0} \frac{1 - \cos(x)}{3x^2}\) (still 0/0) \(\lim_{x \to 0} \frac{\sin(x)}{6x}\) (still 0/0) \(\lim_{x \to 0} \frac{\cos(x)}{6} = \frac{1}{6}\). Source: Wrede and Spiegel (2010), Chapter 4.

Correct Answer: B

Explanation: A positive derivative (\(f'(x) > 0\)) on an interval means that the function is strictly increasing on that interval. This is a direct consequence of the Mean Value Theorem. Source: Adams and Essex (2010), Section 2.8, Theorem 12.