Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 30 Exercises 30.6 Solution: Part aNote that $\cos x\sin x\ge -1$ and $\sin x\ge -1$, we have$$\lim_{x\to\infty}f(x)\ge \lim_{x\to\infty}(x-1)=\infty.$$ Hence…
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 30 Exercises 30.6 Solution: Note that $f(x)=\dfrac{e^xf(x)}{e^x}$, hence we can apply L’Hospital’s Rule \begin{align*} \lim_{x\to\infty}f(x) =&\ \lim_{x\to\infty}\frac{e^xf(x)}{e^x}\\ \text{Apply L’Hospital’s…
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 30 Exercises 30.5 Solution: We start by showing that \begin{equation}\label{30-5-1} \lim_{x\to 0}\left(1+x\right)^{1/x}=e. \end{equation} Indeed, note that $$ \left(1+x\right)^{1/x}=e^{ \ln\left(1+x\right)^{1/x}…
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 30 Exercises 30.3 Solution: Part aSince $|\sin x|\leq 1$, we have\begin{equation}\label{30-3-1}\lim_{x\to \infty}\frac{\sin x}{x}=0.\end{equation} Therefore, by \eqref{30-3-1},$$\lim_{x\to \infty}\frac{x-\sin x}{x}=\lim_{x\to \infty}\frac{1-\frac{\sin…
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 30 Exercises 30.2 Solution: Part aWe can apply L’Hospital’s Rule repeatedly to get the answer. Please check the conditions…
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 30 Exercises 30.1 Solution: Part aWe have $\lim_{x\to 0}(e^{2x}-\cos x)=1-1=0$ and $\lim_{x\to 0} x=0$. Hence we can apply L’Hospital’s…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.10 Solution: (1) We claim that this subset $S$ is an ideal. To that end, suppose $\alpha…
Consider the univariate function $$f(x)=x^3+6x^2-3x-5.$$Find its stationary points and indicate whether they are maximum, minimum, or saddle points. I will assume you are familiar with Calculus, see the book Calculus…
Compute the derivative $f’(x)$ of the function $$f(x)=\exp\left(-\frac{1}{2\sigma^2}(x-\mu)^2\right).$$ Solution: Clearly, we have $$\left(-\frac{1}{2\sigma^2}(x-\mu)^2\right)’=-\frac{1}{2\sigma^2}2(x-\mu)=\frac{-(x-\mu)}{\sigma^2}.$$Therefore, by Chain rule (5.32), we have \begin{align*}f’(x)=&\ \exp\left(-\frac{1}{2\sigma^2}(x-\mu)^2\right)\cdot \left(-\frac{1}{2\sigma^2}(x-\mu)^2\right)’\\=&\ \frac{-(x-\mu)}{\sigma^2}\cdot\exp\left(-\frac{1}{2\sigma^2}(x-\mu)^2\right)\end{align*}
Compute the derivative $f’(x)$ of the logistic sigmoid $$f(x)=\frac{1}{1+\exp(-x)}.$$ Solution: By Chain rule (5.32), we have $$(1+\exp(-x))’=0+\exp(-x)(-x)’=-\exp(-x).$$ By the Quotient rule (5.30), we have\begin{align*}f’(x)=&\ \frac{(1)’(1+\exp(-x))-1(1+\exp(-x))’}{(1+\exp(-x))^2}\\=&\ \frac{0(1+\exp(-x))-(-\exp(-x))}{(1+\exp(-x))^2}\\=&\ \frac{\exp(-x)}{(1+\exp(-x))^2}.\end{align*}