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Intuitively, $\lim_{x \to a} f(x) = L$ means that the value of $f(x)$ gets arbitrarily close to $L$ as $x$ gets sufficiently close to $a$, without actually being equal to $a$. It describes the behavior of the function near the point, not at the point itself.
Source: MT2176 Subject Guide, Section 2.1.2
We say that $\lim_{x \to a} f(x) = L$ if for every number $\epsilon > 0$, there exists a number $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \epsilon$.
Source: Wrede, Chapter 3
The limit $\lim_{x \to a} f(x)$ exists and is equal to $L$ if and only if both the left-hand limit and the right-hand limit exist and are equal to $L$.
$\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = L \text{ and } \lim_{x \to a^+} f(x) = L$
Source: Ostaszewski, Section 17.4
A function $f(x)$ is continuous at $x=c$ if all three of the following conditions are met:
Source: MT2176 Subject Guide, Section 2.2.1
Yes. If a function $f$ is differentiable at a point $c$, then it must be continuous at $c$. Differentiability is a stronger condition than continuity. This is because the existence of the derivative $\lim_{h \to 0} \frac{f(c+h)-f(c)}{h}$ implies that $\lim_{h \to 0} (f(c+h)-f(c)) = 0$, which means $\lim_{x \to c} f(x) = f(c)$.
Source: Wrede, Chapter 4
No. A function can be continuous at a point but not differentiable. A classic example is $f(x) = |x|$ at $x=0$. The function is continuous at $x=0$, but the left-hand and right-hand derivatives are not equal, so the derivative does not exist.
Source: MT2176 Subject Guide, Section 2.2.2
The derivative of a function $f(x)$ with respect to $x$, denoted $f'(x)$, is defined as:
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$
provided this limit exists.
Source: Wrede, Chapter 4
$$ f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} $$
$$ = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} $$
$$ = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x $$
Source: Ostaszewski, Section 17.5
L'Hôpital's Rule can be used to evaluate limits of quotients $\frac{f(x)}{g(x)}$ that result in the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$. If $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ or $\pm\infty$, and $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$.
Source: MT2176 Subject Guide, Section 2.2.4
This limit is of the form $\frac{0}{0}$. Applying L'Hôpital's Rule:
$$ \lim_{x \to 0} \frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{\cos x}{1} = \cos(0) = 1 $$
Source: Ostaszewski, Section 18.8
A jump discontinuity occurs at a point $x=a$ if the left-hand limit $\lim_{x \to a^-} f(x)$ and the right-hand limit $\lim_{x \to a^+} f(x)$ both exist but are not equal. The function "jumps" from one value to another.
Source: Ostaszewski, Section 17.4
An infinite discontinuity occurs at a point $x=a$ if at least one of the one-sided limits of $f(x)$ as $x$ approaches $a$ is infinite. For example, $f(x) = 1/x$ has an infinite discontinuity at $x=0$.
Source: Ostaszewski, Section 17.4
A removable discontinuity occurs at $x=a$ if $\lim_{x \to a} f(x)$ exists but is not equal to $f(a)$. This can be because $f(a)$ is defined to be a different value, or is not defined at all. The discontinuity can be "removed" by redefining $f(a)$ to be equal to the limit.
Source: MT2176 Subject Guide, Section 2.2.1
This is of the form $\infty - \infty$. We rationalize by multiplying by the conjugate: $$ \lim_{x \to \infty} \frac{(\sqrt{x^2+x} - x)(\sqrt{x^2+x} + x)}{\sqrt{x^2+x} + x} = \lim_{x \to \infty} \frac{x^2+x-x^2}{\sqrt{x^2+x} + x} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2(1+1/x)} + x} = \lim_{x \to \infty} \frac{x}{x\sqrt{1+1/x} + x} = \lim_{x \to \infty} \frac{1}{\sqrt{1+1/x} + 1} = \frac{1}{1+1} = \frac{1}{2} $$
Source: Wrede, Chapter 3
If $g(x) \le f(x) \le h(x)$ for all $x$ in an open interval containing $c$ (except possibly at $c$ itself), and if $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$, then $\lim_{x \to c} f(x) = L$. The function $f(x)$ is "squeezed" between $g(x)$ and $h(x)$.
Source: MT2176 Subject Guide, Section 2.1
We know that $-1 \le \sin(1/x) \le 1$ for all $x \neq 0$. Multiplying by $x^2$ (which is non-negative), we get $-x^2 \le x^2 \sin(1/x) \le x^2$. Since $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$, by the Squeeze Theorem, we must have $\lim_{x \to 0} x^2 \sin(1/x) = 0$.
Source: Wrede, Chapter 3
Taylor's Theorem gives an approximation of a $k$-times differentiable function around a given point by a polynomial of degree $k$, called the Taylor polynomial. The theorem provides a formula for the remainder term, which is the difference between the function and its Taylor polynomial. For a function $f(x)$ that is $n+1$ times differentiable on an interval $I$ containing $a$, for any $x$ in $I$, $f(x) = P_n(x) + R_n(x)$, where $P_n(x)$ is the $n$-th order Taylor polynomial and $R_n(x)$ is the remainder.
Source: MT2176 Subject Guide, Section 2.2.3
If $f$ has $n+1$ derivatives in an interval $I$ containing $a$, then for each $x \in I$, the remainder term $R_n(x) = f(x) - P_n(x)$ can be written as: $$ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} $$ for some number $c$ between $a$ and $x$.
Source: Ostaszewski, Section 17.8
This is of the form $1^\infty$. Let $y = (1 + \frac{a}{x})^{bx}$. Then $\ln y = bx \ln(1 + \frac{a}{x})$. This is of the form $\infty \cdot 0$. We rewrite it as $\frac{\ln(1+a/x)}{1/(bx)}$. Applying L'Hôpital's rule: $$ \lim_{x \to \infty} \frac{\frac{1}{1+a/x} \cdot (\frac{-a}{x^2})}{\frac{-1}{bx^2}} = \lim_{x \to \infty} \frac{ab}{1+a/x} = ab $$ So, $\lim_{x \to \infty} y = e^{ab}$.
Source: Wrede, Chapter 3
A function $f$ is of bounded variation on $[a,b]$ if the total variation $V_f[a,b] = \sup_P \sum_{i=1}^n |f(x_i) - f(x_{i-1})|$ is finite, where the supremum is taken over all partitions $P$ of $[a,b]$. Intuitively, it means the function does not "wiggle" infinitely. A function of bounded variation is the difference of two increasing functions.
Source: Ostaszewski, Section 17.16
This is a $\frac{0}{0}$ indeterminate form. Applying L'Hôpital's Rule: $$ \lim_{x \to 0} \frac{\sin(x)}{2x} $$ This is still $\frac{0}{0}$. Applying the rule again: $$ \lim_{x \to 0} \frac{\cos(x)}{2} = \frac{1}{2} $$
Source: MT2176 Subject Guide, Section 2.2.4
As $x$ approaches 5 from the right side (e.g., 5.1, 5.01, 5.001), the denominator $x-5$ is a small positive number. The quotient $\frac{1}{x-5}$ therefore becomes a large positive number. The limit is $\infty$.
Source: Wrede, Chapter 3
As $x$ approaches 5 from the left side (e.g., 4.9, 4.99, 4.999), the denominator $x-5$ is a small negative number. The quotient $\frac{1}{x-5}$ therefore becomes a large negative number. The limit is $-\infty$.
Source: Wrede, Chapter 3
No, the limit does not exist. For the limit to exist, the left-hand limit and the right-hand limit must be equal. In this case, $\lim_{x \to 5^-} \frac{1}{x-5} = -\infty$ and $\lim_{x \to 5^+} \frac{1}{x-5} = +\infty$. Since the one-sided limits are not equal, the two-sided limit does not exist.
Source: Wrede, Chapter 3
If a function $f$ is continuous on a closed interval $[a, b]$, and $N$ is any number between $f(a)$ and $f(b)$ (where $f(a) \neq f(b)$), then there exists at least one number $c$ in the open interval $(a, b)$ such that $f(c) = N$.
Source: Wrede, Chapter 3
The function $f(x) = x^3 - x - 1$ is a polynomial, so it is continuous everywhere. We evaluate the function at the endpoints:
Since $f(1) < 0$ and $f(2) > 0$, and the function is continuous, the Intermediate Value Theorem guarantees there must be a value $c$ in $(1, 2)$ such that $f(c) = 0$.
Source: Wrede, Chapter 3
$$ f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h} $$
Using the identity $\sin(A) - \sin(B) = 2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$:
$$ = \lim_{h \to 0} \frac{2\cos(x+h/2)\sin(h/2)}{h} = \lim_{h \to 0} \cos(x+h/2) \cdot \frac{\sin(h/2)}{h/2} $$
As $h \to 0$, $x+h/2 \to x$ and $h/2 \to 0$. Since $\lim_{u \to 0} \frac{\sin u}{u} = 1$, we get:
$$ = (\cos x)(1) = \cos x $$
Source: Ostaszewski, Section 17.5
As $x$ becomes very large and positive, $-x$ becomes very large and negative. The exponential function $e^y$ approaches 0 as $y$ approaches $-\infty$. Therefore, $\lim_{x \to \infty} e^{-x} = 0$.
Source: MT2176 Subject Guide, Section 2.1.1
This is a $\frac{0}{0}$ form. Apply L'Hôpital's Rule:
$$ \lim_{x \to 0} \frac{\sec^2 x - 1}{3x^2} $$
This is still $\frac{0}{0}$. Apply again:
$$ \lim_{x \to 0} \frac{2\sec x (\sec x \tan x)}{6x} = \lim_{x \to 0} \frac{\sec^2 x \tan x}{3x} $$
This is still $\frac{0}{0}$. Apply again, or use $\lim \frac{\tan x}{x}=1$:
$$ = \lim_{x \to 0} \frac{\sec^2 x}{3} \cdot \lim_{x \to 0} \frac{\tan x}{x} = \frac{1}{3} \cdot 1 = \frac{1}{3} $$
Source: Ostaszewski, Section 18.8
The first-order Taylor polynomial is $P_1(x) = f(0) + f'(0)(x-0)$.
$f(x) = \ln(1+x) \implies f(0) = \ln(1) = 0$.
$f'(x) = \frac{1}{1+x} \implies f'(0) = 1$.
Therefore, $P_1(x) = 0 + 1(x) = x$. This is the linear approximation of $\ln(1+x)$ near $x=0$.
Source: MT2176 Subject Guide, Section 2.2.3
We are using the first-order Taylor polynomial $P_1(x)=x$ for $f(x)=\sin x$ around $a=0$. The error is given by the remainder term $R_1(x) = \frac{f\"(c)}{2!}x^2$.
$f\"(x) = -\sin x$. So, $R_1(0.1) = \frac{-\sin c}{2}(0.1)^2$ for some $c \in (0, 0.1)$.
Since $|-\sin c| \le 1$, the error is bounded by: $$ |R_1(0.1)| \le \frac{1}{2}(0.01) = 0.005 $$
Source: MT2176 Subject Guide, Section 2.2.3
This is a $\frac{0}{0}$ indeterminate form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 1} \frac{\frac{d}{dx}(x^n - 1)}{\frac{d}{dx}(x - 1)} = \lim_{x \to 1} \frac{nx^{n-1}}{1} = n(1)^{n-1} = n $$
Alternatively, one can factor the numerator as $(x-1)(x^{n-1} + x^{n-2} + \dots + x + 1)$ and cancel the $(x-1)$ term.
Source: Ostaszewski, Section 18.8
No, the function is not continuous at $x=2$ because it is not defined at that point (division by zero). However, this is a removable discontinuity. We can simplify the function for $x \neq 2$: $$ f(x) = \frac{(x-2)(x+2)}{x-2} = x+2 $$ The limit as $x \to 2$ is $\lim_{x \to 2} (x+2) = 4$. If we define $f(2)=4$, the function becomes continuous.
Source: MT2176 Subject Guide, Section 2.2.1
$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h} $$
$$ = \lim_{h \to 0} \frac{-h}{hx(x+h)} = \lim_{h \to 0} \frac{-1}{x(x+h)} = -\frac{1}{x^2} $$
Source: MT2176 Subject Guide, Section 2.2.2
This is of the form $\infty \cdot 0$. Let $t = 1/x$. As $x \to \infty$, $t \to 0^+$. The limit becomes:
$$ \lim_{t \to 0^+} \frac{\sin t}{t} $$
This is a standard limit which equals 1. We can also confirm this with L'Hôpital's Rule.
Source: Wrede, Chapter 3
The second-order Taylor polynomial is $P_2(x) = f(0) + f'(0)x + \frac{f\"(0)}{2!}x^2$.
For $f(x) = e^x$, all derivatives are $e^x$. So, $f(0)=1$, $f'(0)=1$, and $f\"(0)=1$.
Therefore, $P_2(x) = 1 + 1 \cdot x + \frac{1}{2}x^2 = 1 + x + \frac{x^2}{2}$.
Source: MT2176 Subject Guide, Section 2.2.3
This is an indeterminate form of type $0 \cdot (-\infty)$. We rewrite it as a quotient to apply L'Hôpital's Rule:
$$ \lim_{x \to 0^+} \frac{\ln x}{1/x} $$
This is now of the form $\frac{-\infty}{\infty}$. Applying the rule:
$$ \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0 $$
Source: Ostaszewski, Section 18.8
A sequence is an ordered list of numbers, such as $u_1, u_2, u_3, \dots$.
A series is the sum of the terms of a sequence, such as $u_1 + u_2 + u_3 + \dots$. The value of an infinite series is defined as the limit of its sequence of partial sums, if that limit exists.
Source: Wrede, Chapter 2
Divide the numerator and denominator by the highest power of $x$, which is $x^3$:
$$ \lim_{x \to \infty} \frac{5 - 2/x^2 + 1/x^3}{2 + 1/x - 9/x^3} $$
As $x \to \infty$, terms like $1/x$, $1/x^2$, and $1/x^3$ go to 0. So the limit is:
$$ \frac{5 - 0 + 0}{2 + 0 - 0} = \frac{5}{2} $$
Source: MT2176 Subject Guide, Section 2.1.1
The derivative of a constant function is zero. Using the limit definition:
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{c - c}{h} = \lim_{h \to 0} \frac{0}{h} = 0 $$
Source: Wrede, Chapter 4
This is a fundamental limit that defines the number $e$.
$$ \lim_{x \to 0} (1+x)^{1/x} = e $$
To show this, let $y = (1+x)^{1/x}$. Then $\ln y = \frac{\ln(1+x)}{x}$. We know from L'Hôpital's rule or Taylor series that $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$. Since the natural logarithm is a continuous function, $\lim_{x \to 0} \ln y = \ln(\lim_{x \to 0} y) = 1$, which implies $\lim_{x \to 0} y = e^1 = e$.
Source: Wrede, Chapter 3
$P_3(x) = f(0) + f'(0)x + \frac{f\"(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3$.
Therefore, $P_3(x) = 1 + 0\cdot x + \frac{-1}{2}x^2 + 0 \cdot x^3 = 1 - \frac{x^2}{2}$.
Source: MT2176 Subject Guide, Section 2.2.3
This is a $\frac{0}{0}$ indeterminate form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 1} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x-1)} = \lim_{x \to 1} \frac{1/x}{1} = 1 $$
Source: Ostaszewski, Section 18.8
We check the one-sided limits:
The value of the function at the point is $f(1) = 1+1 = 2$. Since the left-hand limit, right-hand limit, and function value are all equal, the function is continuous at $x=1$.
Source: MT2176 Subject Guide, Section 2.2.1
$$ f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h} $$
$$ = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} $$
$$ = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2 $$
Source: Ostaszewski, Section 17.5
This is an $\frac{\infty}{\infty}$ indeterminate form. Applying L'Hôpital's Rule:
$$ \lim_{x \to \infty} \frac{e^x}{2x} $$
This is still $\frac{\infty}{\infty}$. Applying the rule again:
$$ \lim_{x \to \infty} \frac{e^x}{2} = \infty $$
The limit is infinite, meaning the function grows without bound.
Source: Ostaszewski, Section 18.8
A number $L$ is the limit of an infinite sequence {$a_n$} if for any positive number $\epsilon$, there is a positive integer $N$ such that for all $n > N$, we have $|a_n - L| < \epsilon$. This means the terms of the sequence get arbitrarily close to $L$ as $n$ gets large.
Source: Wrede, Chapter 2
We can rewrite the term as $a_n = 1 + \frac{1}{n}$. As $n \to \infty$, the term $\frac{1}{n} \to 0$.
Therefore, $\lim_{n \to \infty} a_n = \lim_{n \to \infty} (1 + \frac{1}{n}) = 1 + 0 = 1$.
Source: Wrede, Chapter 2
We must check the one-sided limits.
Since the left-hand and right-hand limits are not equal, the limit does not exist.
Source: Wrede, Chapter 3
The Taylor series for $\ln(1+x)$ about $x=0$ is $x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$.
$$ \lim_{x \to 0} \frac{(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots) - x}{x^2} = \lim_{x \to 0} \frac{-\frac{x^2}{2} + \frac{x^3}{3} - \dots}{x^2} $$
$$ = \lim_{x \to 0} (-\frac{1}{2} + \frac{x}{3} - \dots) = -\frac{1}{2} $$
Source: MT2176 Subject Guide, Section 2.2.3
A proper integral, $\int_a^b f(x) dx$, has a finite interval of integration $[a, b]$ and an integrand $f(x)$ that is bounded on that interval.
An improper integral has either an infinite interval of integration (e.g., $[a, \infty)$) or an integrand with a finite number of infinite discontinuities within the interval of integration.
Source: Ostaszewski, Section 18.6
This is an indeterminate form $0^0$. Let $y = x^x$, so $\ln y = x \ln x$. We have already shown that $\lim_{x \to 0^+} x \ln x = 0$.
Therefore, $\lim_{x \to 0^+} \ln y = 0$. Since the exponential function is continuous, this means $\lim_{x \to 0^+} y = e^0 = 1$.
Source: Ostaszewski, Section 18.8
The derivative of a function $f$ at a point $x=c$, denoted $f'(c)$, represents the slope of the tangent line to the graph of $y=f(x)$ at the point $(c, f(c))$. It is the instantaneous rate of change of the function at that point.
Source: Wrede, Chapter 4
$$ f'(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos(x)}{h} $$
Using the identity $\cos(A) - \cos(B) = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$:
$$ = \lim_{h \to 0} \frac{-2\sin(x+h/2)\sin(h/2)}{h} = -\lim_{h \to 0} \sin(x+h/2) \cdot \frac{\sin(h/2)}{h/2} $$
As $h \to 0$, this becomes $(-\sin x)(1) = -\sin x$.
Source: Ostaszewski, Section 17.5
This is of the form $1^\infty$. We can rewrite the expression: $$ \lim_{x \to \infty} (\frac{x+1-1}{x+1})^x = \lim_{x \to \infty} (1 - \frac{1}{x+1})^x $$ Let $y = x+1$. As $x \to \infty$, $y \to \infty$. The limit becomes: $$ \lim_{y \to \infty} (1 - \frac{1}{y})^{y-1} = \lim_{y \to \infty} (1 - \frac{1}{y})^y \cdot \lim_{y \to \infty} (1 - \frac{1}{y})^{-1} $$ Using the standard limit $\lim_{n \to \infty} (1+a/n)^n = e^a$, the first part is $e^{-1}$. The second part is $1^{-1}=1$. So the limit is $e^{-1}$.
Source: Wrede, Chapter 3
$P_1(x) = f(0) + f'(0)x$.
$f(x) = (1+x)^{1/2} \implies f(0) = 1$.
$f'(x) = \frac{1}{2}(1+x)^{-1/2} \implies f'(0) = \frac{1}{2}$.
Therefore, $P_1(x) = 1 + \frac{1}{2}x$. This is the linear approximation, also known as the binomial approximation for this case.
Source: MT2176 Subject Guide, Section 2.2.3
This is a $\frac{0}{0}$ indeterminate form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 0} \frac{e^x - (-e^{-x})}{1} = \lim_{x \to 0} (e^x + e^{-x}) = e^0 + e^0 = 1 + 1 = 2 $$
Source: Ostaszewski, Section 18.8
This can be written as $a_n = (1 - \frac{1}{n})^n$. This is a standard limit form.
$$ \lim_{n \to \infty} (1 - \frac{1}{n})^n = e^{-1} = \frac{1}{e} $$
Source: Wrede, Chapter 2
No. The function is not defined at $x=0$. Furthermore, the limit $\lim_{x \to 0} \sin(1/x)$ does not exist. As $x$ approaches 0, $1/x$ approaches infinity, and the sine function oscillates infinitely between -1 and 1 without approaching a single value. Therefore, it has a non-removable discontinuity at $x=0$.
Source: Ostaszewski, Section 17.4
$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{2(x+h)+1} - \sqrt{2x+1}}{h} $$
Multiply by the conjugate:
$$ = \lim_{h \to 0} \frac{(2x+2h+1) - (2x+1)}{h(\sqrt{2x+2h+1} + \sqrt{2x+1})} = \lim_{h \to 0} \frac{2h}{h(\dots)} $$
$$ = \lim_{h \to 0} \frac{2}{\sqrt{2x+2h+1} + \sqrt{2x+1}} = \frac{2}{2\sqrt{2x+1}} = \frac{1}{\sqrt{2x+1}} $$
Source: Ostaszewski, Section 17.5
This is a $\frac{0}{0}$ form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 1} \frac{3x^2 - 4x}{3x^2} = \frac{3(1)^2 - 4(1)}{3(1)^2} = \frac{-1}{3} $$
Source: Ostaszewski, Section 18.8
The Taylor series for $e^x$ about $x=0$ is $1 + x + \frac{x^2}{2!} + \dots$.
$$ \lim_{x \to 0} \frac{(1 + x + \frac{x^2}{2!} + \dots) - 1}{x} = \lim_{x \to 0} \frac{x + \frac{x^2}{2} + \dots}{x} $$
$$ = \lim_{x \to 0} (1 + \frac{x}{2} + \dots) = 1 $$
Source: MT2176 Subject Guide, Section 2.2.3
Every bounded, monotonic sequence converges to a limit. If the sequence is monotonically increasing, it converges to its least upper bound. If it is monotonically decreasing, it converges to its greatest lower bound.
Source: Wrede, Chapter 2
This is an $\frac{\infty}{\infty}$ form. Applying L'Hôpital's Rule:
$$ \lim_{x \to \infty} \frac{\frac{2x}{x^2+1}}{\frac{3x^2}{x^3+1}} = \lim_{x \to \infty} \frac{2x(x^3+1)}{3x^2(x^2+1)} = \lim_{x \to \infty} \frac{2x^4+2x}{3x^4+3x^2} $$
Divide by $x^4$: $$ \lim_{x \to \infty} \frac{2+2/x^3}{3+3/x^2} = \frac{2}{3} $$
Source: Ostaszewski, Section 18.8
A function is continuous on an open interval $(a,b)$ if it is continuous at every point in the interval.
For a closed interval $[a,b]$, the function must be continuous on $(a,b)$, and it must also be continuous from the right at $a$ (i.e., $\lim_{x \to a^+} f(x) = f(a)$) and continuous from the left at $b$ (i.e., $\lim_{x \to b^-} f(x) = f(b)$).
Source: Wrede, Chapter 3
$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h+1} - \frac{1}{x+1}}{h} $$
$$ = \lim_{h \to 0} \frac{x+1 - (x+h+1)}{h(x+1)(x+h+1)} = \lim_{h \to 0} \frac{-h}{h(x+1)(x+h+1)} $$
$$ = \lim_{h \to 0} \frac{-1}{(x+1)(x+h+1)} = -\frac{1}{(x+1)^2} $$
Source: MT2176 Subject Guide, Section 2.2.2
This is a $\frac{0}{0}$ form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 0} \frac{a\cos(ax)}{b\cos(bx)} = \frac{a\cos(0)}{b\cos(0)} = \frac{a}{b} $$
Source: Ostaszewski, Section 18.8
$P_2(x) = f(0) + f'(0)x + \frac{f\"(0)}{2!}x^2$.
Therefore, $P_2(x) = 0 + 1 \cdot x + \frac{0}{2}x^2 = x$.
Source: MT2176 Subject Guide, Section 2.2.3
This is an $\infty^0$ indeterminate form. Let $y = x^{1/x}$, so $\ln y = \frac{\ln x}{x}$.
We have already shown that $\lim_{x \to \infty} \frac{\ln x}{x} = 0$ using L'Hôpital's Rule.
Therefore, $\lim_{x \to \infty} \ln y = 0$. Since the exponential function is continuous, this means $\lim_{x \to \infty} y = e^0 = 1$.
Source: Ostaszewski, Section 18.8
The sequence is $-1, 1, -1, 1, \dots$. The terms oscillate between -1 and 1 and do not approach a single value. Therefore, the limit does not exist.
Source: Wrede, Chapter 2
No. The function is not defined at $x=0$. Furthermore, as $x$ approaches 0 from either the left or the right, $f(x)$ increases without bound. $$ \lim_{x \to 0} \frac{1}{x^2} = \infty $$ This is an infinite discontinuity.
Source: Ostaszewski, Section 17.4
$$ f'(x) = \lim_{h \to 0} \frac{(5(x+h)+3) - (5x+3)}{h} $$
$$ = \lim_{h \to 0} \frac{5x+5h+3 - 5x-3}{h} = \lim_{h \to 0} \frac{5h}{h} = \lim_{h \to 0} 5 = 5 $$
Source: MT2176 Subject Guide, Section 2.2.2
This is a $\frac{0}{0}$ indeterminate form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 0} \frac{\frac{1}{\sqrt{1-x^2}}}{1} = \frac{1}{\sqrt{1-0}} = 1 $$
Source: Ostaszewski, Section 18.8
The Taylor series for $\cos x$ about $x=0$ is $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$.
$$ \lim_{x \to 0} \frac{1 - (1 - \frac{x^2}{2!} + \dots)}{x} = \lim_{x \to 0} \frac{\frac{x^2}{2} - \frac{x^4}{24} + \dots}{x} $$
$$ = \lim_{x \to 0} (\frac{x}{2} - \frac{x^3}{24} + \dots) = 0 $$
Source: MT2176 Subject Guide, Section 2.2.3
We can use the Squeeze Theorem. We know that $-1 \le \sin n \le 1$ for all $n$.
Therefore, $-\frac{1}{n} \le \frac{\sin n}{n} \le \frac{1}{n}$.
Since $\lim_{n \to \infty} (-\frac{1}{n}) = 0$ and $\lim_{n \to \infty} \frac{1}{n} = 0$, the Squeeze Theorem tells us that $\lim_{n \to \infty} \frac{\sin n}{n} = 0$.
Source: Wrede, Chapter 2
This is of the form $\infty - \infty$. We rationalize by multiplying by the conjugate:
$$ \lim_{x \to \infty} \frac{(x^2+1) - (x^2-1)}{\sqrt{x^2+1} + \sqrt{x^2-1}} = \lim_{x \to \infty} \frac{2}{\sqrt{x^2+1} + \sqrt{x^2-1}} $$
As $x \to \infty$, the denominator approaches $\infty$. Therefore, the limit is 0.
Source: Wrede, Chapter 3
No. The floor function, which gives the greatest integer less than or equal to $x$, has a jump discontinuity at every integer. For example, at $x=2$:
Since the one-sided limits are not equal, the function is not continuous at any integer.
Source: Ostaszewski, Section 17.13
$$ f'(x) = \lim_{h \to 0} \frac{[2(x+h)^2 - 3(x+h)] - [2x^2 - 3x]}{h} $$
$$ = \lim_{h \to 0} \frac{2x^2+4xh+2h^2-3x-3h-2x^2+3x}{h} $$
$$ = \lim_{h \to 0} \frac{4xh+2h^2-3h}{h} = \lim_{h \to 0} (4x+2h-3) = 4x-3 $$
Source: MT2176 Subject Guide, Section 2.2.2
Using the identity $1 - \cos^2 x = \sin^2 x$, the limit becomes:
$$ \lim_{x \to 0} \frac{x^2}{\sin^2 x} = \lim_{x \to 0} (\frac{x}{\sin x})^2 $$
Since we know $\lim_{x \to 0} \frac{\sin x}{x} = 1$, it follows that $\lim_{x \to 0} \frac{x}{\sin x} = 1$. Therefore, the limit is $1^2 = 1$.
Source: Ostaszewski, Section 18.8
$P_4(x) = f(0) + f'(0)x + \frac{f\"(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4$.
The derivatives are $\cos x, -\sin x, -\cos x, \sin x, \cos x$. At $x=0$, these are $1, 0, -1, 0, 1$.
$$ P_4(x) = 1 + 0 - \frac{1}{2}x^2 + 0 + \frac{1}{24}x^4 = 1 - \frac{x^2}{2} + \frac{x^4}{24} $$
Source: MT2176 Subject Guide, Section 2.2.3
This is of the form $\infty - \infty$. We can factor out the dominant term, $x$:
$$ \lim_{x \to \infty} x(1 - \frac{\ln x}{x}) $$
We know $\lim_{x \to \infty} \frac{\ln x}{x} = 0$. So the expression inside the parenthesis approaches 1. Since the $x$ outside approaches $\infty$, the overall limit is $\infty$.
Source: Wrede, Chapter 3
This is an indeterminate form $\infty^0$. Let $y = n^{1/n}$, so $\ln y = \frac{\ln n}{n}$.
Using L'Hôpital's Rule on the continuous version $\frac{\ln x}{x}$:
$$ \lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0 $$
Therefore, $\lim_{n \to \infty} \ln a_n = 0$, which implies $\lim_{n \to \infty} a_n = e^0 = 1$.
Source: Ostaszewski, Section 18.7
No. For a function to be differentiable at a point, it must first be continuous at that point. The function $f(x) = \frac{1}{x-2}$ is not continuous at $x=2$ because it is not defined there (it has an infinite discontinuity). Therefore, it cannot be differentiable at $x=2$.
Source: Wrede, Chapter 4
$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \lim_{h \to 0} \frac{x^2 - (x+h)^2}{hx^2(x+h)^2} $$
$$ = \lim_{h \to 0} \frac{x^2 - (x^2+2xh+h^2)}{hx^2(x+h)^2} = \lim_{h \to 0} \frac{-2xh-h^2}{hx^2(x+h)^2} $$
$$ = \lim_{h \to 0} \frac{-2x-h}{x^2(x+h)^2} = \frac{-2x}{x^2(x^2)} = -\frac{2}{x^3} $$
Source: MT2176 Subject Guide, Section 2.2.2
This is a $\frac{0}{0}$ indeterminate form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 0} \frac{\sec^2 x}{1} = \sec^2(0) = 1^2 = 1 $$
Source: Ostaszewski, Section 18.8
The Taylor series for $\sin x$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$.
$$ \lim_{x \to 0} \frac{x - (x - \frac{x^3}{6} + \frac{x^5}{120} - \dots)}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{6} - \frac{x^5}{120} + \dots}{x^3} $$
$$ = \lim_{x \to 0} (\frac{1}{6} - \frac{x^2}{120} + \dots) = \frac{1}{6} $$
Source: MT2176 Subject Guide, Section 2.2.3
The sequence is $\cos(\pi), \cos(2\pi), \cos(3\pi), \dots$, which evaluates to $-1, 1, -1, 1, \dots$. Since the terms oscillate between -1 and 1 and do not approach a single value, the limit does not exist.
Source: Wrede, Chapter 2
Using logarithm properties, this is $\lim_{x \to \infty} \ln(\frac{x+1}{x}) = \lim_{x \to \infty} \ln(1 + \frac{1}{x})$.
As $x \to \infty$, $\frac{1}{x} \to 0$, so $1 + \frac{1}{x} \to 1$.
Since $\ln$ is a continuous function, we can take the limit inside: $\ln(\lim_{x \to \infty} (1 + \frac{1}{x})) = \ln(1) = 0$.
Source: Wrede, Chapter 3
The function $f(x) = \sqrt{x}$ is defined for $x \ge 0$. At $x=0$, we can only consider the right-hand limit.
$\lim_{x \to 0^+} \sqrt{x} = 0$.
The value of the function is $f(0) = \sqrt{0} = 0$.
Since the right-hand limit equals the function value, the function is continuous from the right at $x=0$. It is considered continuous on its domain $[0, \infty)$.
Source: Wrede, Chapter 3
$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}} $$
Rationalize the numerator:
$$ = \lim_{h \to 0} \frac{x - (x+h)}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} = \lim_{h \to 0} \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} $$
$$ = \frac{-1}{\sqrt{x}\sqrt{x}(2\sqrt{x})} = -\frac{1}{2x^{3/2}} $$
Source: MT2176 Subject Guide, Section 2.2.2
This is a $\frac{0}{0}$ form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 0} \frac{2x\cos(x^2)}{1} = 2(0)\cos(0) = 0 $$
Alternatively, rewrite as $\lim_{x \to 0} x \cdot \frac{\sin(x^2)}{x^2}$. The first part goes to 0 and the second part goes to 1, so the product is 0.
Source: Ostaszewski, Section 18.8
$P_3(x) = f(0) + f'(0)x + \frac{f\"(0)}{2}x^2 + \frac{f'''(0)}{6}x^3$.
$$ P_3(x) = 1 + 2x + \frac{4}{2}x^2 + \frac{8}{6}x^3 = 1 + 2x + 2x^2 + \frac{4}{3}x^3 $$
Source: MT2176 Subject Guide, Section 2.2.3
This is a $1^\infty$ form. We rewrite the base: $$ \frac{x+3}{x-1} = \frac{x-1+4}{x-1} = 1 + \frac{4}{x-1} $$ Let $y = x-1$. As $x \to \infty$, $y \to \infty$. The limit becomes: $$ \lim_{y \to \infty} (1 + \frac{4}{y})^{y+1} = \lim_{y \to \infty} (1 + \frac{4}{y})^y \cdot \lim_{y \to \infty} (1 + \frac{4}{y})^1 $$ The first limit is $e^4$ and the second is 1. The result is $e^4$.
Source: Wrede, Chapter 3
Consider the ratio of consecutive terms: $$ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{n+1}{(n+1)^{n+1}} \cdot n^n = \frac{n^n}{(n+1)^n} = (\frac{n}{n+1})^n = (1 - \frac{1}{n+1})^n $$ As $n \to \infty$, this limit approaches $e^{-1} < 1$. Since the ratio of consecutive terms is less than 1, the terms are decreasing and must approach 0. Therefore, $\lim_{n \to \infty} a_n = 0$.
Source: Wrede, Chapter 2
We check the conditions for continuity.
The function is not continuous at $x=0$. It has a removable discontinuity.
Source: MT2176 Subject Guide, Section 2.2.1
$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{2(x+h)-1} - \frac{1}{2x-1}}{h} $$
$$ = \lim_{h \to 0} \frac{(2x-1) - (2x+2h-1)}{h(2x-1)(2x+2h-1)} = \lim_{h \to 0} \frac{-2h}{h(2x-1)(2x+2h-1)} $$
$$ = \lim_{h \to 0} \frac{-2}{(2x-1)(2x+2h-1)} = -\frac{2}{(2x-1)^2} $$
Source: MT2176 Subject Guide, Section 2.2.2
This is a $\frac{0}{0}$ form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 0} \frac{\frac{2}{1+(2x)^2}}{3\cos(3x)} = \frac{\frac{2}{1+0}}{3\cos(0)} = \frac{2}{3} $$
Source: Ostaszewski, Section 18.8
The Taylor series for $\ln(1-x)$ is $-x - \frac{x^2}{2} - \frac{x^3}{3} - \dots$.
$$ \lim_{x \to 0} \frac{(-x - \frac{x^2}{2} - \dots) + x}{x^2} = \lim_{x \to 0} \frac{-\frac{x^2}{2} - \frac{x^3}{3} - \dots}{x^2} $$
$$ = \lim_{x \to 0} (-\frac{1}{2} - \frac{x}{3} - \dots) = -\frac{1}{2} $$
Source: MT2176 Subject Guide, Section 2.2.3
We use the ratio test for sequences. Consider the ratio of consecutive terms:
$$ \frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} = \frac{2}{n+1} $$
As $n \to \infty$, this ratio $\frac{2}{n+1} \to 0$. Since the limit of the ratio is less than 1, the sequence converges to 0. Therefore, $\lim_{n \to \infty} a_n = 0$.
Source: Wrede, Chapter 2
This is a $0 \cdot \infty$ indeterminate form. We rewrite it as a quotient:
$$ \lim_{x \to 1} \frac{x-1}{\cot(\frac{\pi x}{2})} $$
This is now a $\frac{0}{0}$ form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 1} \frac{1}{-\csc^2(\frac{\pi x}{2}) \cdot \frac{\pi}{2}} = \frac{1}{-\frac{\pi}{2} (1)^2} = -\frac{2}{\pi} $$
Source: Ostaszewski, Section 18.8
We check the one-sided limits.
Since the left-hand and right-hand limits are not equal, the limit does not exist.
Source: Wrede, Chapter 3
We use the limit definition of the derivative:
$$ f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{h|h| - 0}{h} = \lim_{h \to 0} |h| = 0 $$
The derivative exists and is equal to 0. Note that for $x>0$, $f(x)=x^2$ and $f'(x)=2x$. For $x<0$, $f(x)=-x^2$ and $f'(x)=-2x$. Both approach 0 as $x \to 0$.
Source: MT2176 Subject Guide, Section 2.2.2
This is an $\infty - \infty$ form. We combine the terms:
$$ \lim_{x \to 0} (\frac{1}{\sin x} - \frac{\cos x}{\sin x}) = \lim_{x \to 0} \frac{1 - \cos x}{\sin x} $$
This is now a $\frac{0}{0}$ form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 0} \frac{\sin x}{\cos x} = \frac{0}{1} = 0 $$
Source: Ostaszewski, Section 18.8
The Taylor series for $\cos u$ is $1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots$. Let $u=x^2$.
$$ \lim_{x \to 0} \frac{1 - (1 - \frac{(x^2)^2}{2!} + \frac{(x^2)^4}{4!} - \dots)}{x^4} = \lim_{x \to 0} \frac{\frac{x^4}{2} - \frac{x^8}{24} + \dots}{x^4} $$
$$ = \lim_{x \to 0} (\frac{1}{2} - \frac{x^4}{24} + \dots) = \frac{1}{2} $$
Source: MT2176 Subject Guide, Section 2.2.3
Divide the numerator and denominator by the highest power of $n$, which is $n^2$:
$$ \lim_{n \to \infty} \frac{1 + 1/n^2}{2 - 1/n^2} = \frac{1+0}{2-0} = \frac{1}{2} $$
Source: Wrede, Chapter 2
First, combine the terms: $$ \lim_{x \to \infty} \frac{x^2+1 - (ax+b)(x+1)}{x+1} = \lim_{x \to \infty} \frac{x^2+1 - (ax^2 + (a+b)x + b)}{x+1} = \lim_{x \to \infty} \frac{(1-a)x^2 - (a+b)x + (1-b)}{x+1} $$ For the limit to be 0, the degree of the numerator must be less than the degree of the denominator. This requires the coefficients of $x^2$ and $x$ to be zero.
$1-a=0 \implies a=1$.
$-(a+b)=0 \implies -(1+b)=0 \implies b=-1$.
Source: Wrede, Chapter 3
We check the limit definition of the derivative at $x=0$:
$$ f'(0) = \lim_{h \to 0} \frac{h\sin(1/h) - 0}{h} = \lim_{h \to 0} \sin(1/h) $$
This limit does not exist because $\sin(1/h)$ oscillates between -1 and 1 as $h \to 0$. Therefore, the function is not differentiable at $x=0$, even though it is continuous there.
Source: Ostaszewski, Section 17.4
This is very difficult to do directly from the definition. The limit definition is primarily for understanding the concept and proving basic rules. For this function, we would use the chain rule: $f'(x) = e^{x^2} \cdot \frac{d}{dx}(x^2) = 2x e^{x^2}$. The question highlights the importance of differentiation rules over first principles for complex functions.
Source: MT2176 Subject Guide, Section 2.2.2
This is a $\frac{0}{0}$ form. Let $u = x^2$. As $x \to 0$, $u \to 0$. The limit becomes $\lim_{u \to 0} \frac{1 - e^{-u}}{u}$.
Applying L'Hôpital's Rule:
$$ \lim_{u \to 0} \frac{-(-e^{-u})}{1} = \lim_{u \to 0} e^{-u} = e^0 = 1 $$
Source: Ostaszewski, Section 18.8
The Taylor series for $\arctan x$ is $x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$.
$$ \lim_{x \to 0} \frac{x - (x - \frac{x^3}{3} + \frac{x^5}{5} - \dots)}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{3} - \frac{x^5}{5} + \dots}{x^3} $$
$$ = \lim_{x \to 0} (\frac{1}{3} - \frac{x^2}{5} + \dots) = \frac{1}{3} $$
Source: MT2176 Subject Guide, Section 2.2.3
Consider the ratio of consecutive terms: $$ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{n+1}{(n+1)^{n+1}} \cdot n^n = \frac{n^n}{(n+1)^n} = (\frac{n}{n+1})^n = (1 - \frac{1}{n+1})^n $$ As $n \to \infty$, this limit approaches $e^{-1} < 1$. Since the ratio of consecutive terms is less than 1, the terms are decreasing and must approach 0. Therefore, $\lim_{n \to \infty} a_n = 0$.
Source: Wrede, Chapter 2
This is an $\infty - \infty$ form. Combine the fractions:
$$ \lim_{x \to 0} \frac{e^x - 1 - x}{x(e^x-1)} $$
This is a $\frac{0}{0}$ form. Apply L'Hôpital's Rule:
$$ \lim_{x \to 0} \frac{e^x - 1}{e^x-1 + xe^x} $$
Still $\frac{0}{0}$. Apply again:
$$ \lim_{x \to 0} \frac{e^x}{e^x + e^x + xe^x} = \frac{1}{1+1+0} = \frac{1}{2} $$
Source: Ostaszewski, Section 18.8
For $x \neq 0$, we can write this as $f(x) = |x| \cdot \frac{|x|}{|x|} = |x|$.
So we need to find $\lim_{x \to 0} |x|$.
$$ \lim_{x \to 0} |x| = 0 $$
The limit exists and is 0.
Source: Wrede, Chapter 3
This is very difficult to do directly from the definition. The chain rule is the appropriate tool here: $f'(x) = -\sin(x^2) \cdot \frac{d}{dx}(x^2) = -2x \sin(x^2)$. This question again emphasizes the power and necessity of differentiation rules over using first principles for most functions.
Source: MT2176 Subject Guide, Section 2.2.2
This is a $\frac{0}{0}$ form. Applying L'Hôpital's Rule:
$$ \lim_{x \to 0} \frac{3x^2}{1 - \cos x} $$
Still $\frac{0}{0}$. Apply again:
$$ \lim_{x \to 0} \frac{6x}{\sin x} = 6 \lim_{x \to 0} \frac{x}{\sin x} = 6(1) = 6 $$
Source: Ostaszewski, Section 18.8
The Taylor series for $\ln(1+x)$ is $x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$.
$$ \lim_{x \to 0} \frac{x(x - \frac{x^2}{2} + \dots) - x^2}{x^3} = \lim_{x \to 0} \frac{x^2 - \frac{x^3}{2} - x^2 + \dots}{x^3} $$
$$ = \lim_{x \to 0} \frac{-\frac{x^3}{2} + \dots}{x^3} = -\frac{1}{2} $$
Source: MT2176 Subject Guide, Section 2.2.3
Consider the continuous function $f(x) = \frac{\ln x}{\sqrt{x}}$. This is an $\frac{\infty}{\infty}$ form as $x \to \infty$.
Applying L'Hôpital's Rule:
$$ \lim_{x \to \infty} \frac{1/x}{1/(2\sqrt{x})} = \lim_{x \to \infty} \frac{2\sqrt{x}}{x} = \lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0 $$
Therefore, the sequence converges to 0.
Source: Wrede, Chapter 2
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