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 $$ \left(1+x\right)^{1/x}=e^{ \ln\left(1+x\right)^{1/x}…
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 30 Exercises 30.4 Solution: First, we show that $\lim_{x\to 0^+}f(x)$ exists if $\lim_{y\to \infty}g(y)$ exists. Suppose $\lim_{y\to \infty}g(y)=L$. For…
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 30 Exercises 30.3 Solution: Part aSince $|\sin x|\leq 1$, we have\begin{equation}\label{30-3-1}\lim_{x\to \infty}\frac{\sin x}{x}=0.\end{equation} Therefore, by \eqref{30-3-1},$$\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 Elementary Analysis: The Theory of Calculus Second Edition Section 9 Exercise 9.8 Solution: Part aWe have $\lim n^3=\infty$. Part bWe have $\lim(-n^3)=-\infty$. Part cWe have $\lim(-n)^n$ does not…
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 9 Exercise 9.7 Solution: We have seen that $0\le s_n<\sqrt{\dfrac{2}{n-1}}$. By Sequeeze-Theorem/Exercise 8.5, it suffices to show that $\lim…
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 9 Exercise 9.6 Solution: Part a This part is the same as that of Exercise 9.4 and Exercise 9.5.…