Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 30 Exercises 30.3
Solution:
Part a
Since $|\sin x|\leq 1$, 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},
$$
\lim_{x\to \infty}\frac{x-\sin x}{x}=\lim_{x\to \infty}\frac{1-\frac{\sin x}{x}}{1}=\frac{1-0}{1}=1.
$$
In this case, we cannot apply L’Hospital’s Rule because the limit $\lim_{x\to \infty}(1-\cos x)$ does not exists.
Part b
We consider
$$
\ln x^{\sin(1/x)}=\sin(1/x)\ln x=-\sin (1/x)\ln(1/x).
$$ Using substitution $u=1/x$, we have
\begin{align*}
\lim_{x\to \infty }\ln x^{\sin(1/x)}
=&\ \lim_{u\to 0^+ }-\sin u\ln u\\
=&\ -\lim_{u\to 0^+ }\frac{\ln u}{\csc u}\\
\text{L’Hospital’s Rule}\ =&\ \lim_{u\to 0^+ }\frac{\frac{1}{u}}{\csc u\cot u}\\
=&\ \lim_{u\to 0^+ }\frac{\sin^2 u}{u\cos u}\\
=&\ \lim_{u\to 0^+ }\frac{\sin^2 u}{u^2}\lim_{u\to 0^+ }\frac{u}{\cos u}\\
\text{Use Example 1}\ =&\ 1\cdot \frac{0}{1}=0.
\end{align*} Here we used Example 1,
$$
\lim_{x\to 0}\frac{\sin x}{x}=1.
$$ Therefore by Theorem 20.5, we have
$$
\lim_{x\to \infty} x^{\sin(1/x)}=e^{0}=1.
$$
Part c
Clearly, we have
$$
\lim_{x\to 0^+}(1+\cos x)=2
$$ and
$$
\lim_{x\to 0^+}(e^x-1)=(1^+-1)=0^+.
$$ Therefore,
$$
\lim_{x\to 0^+}\frac{1+\cos x}{e^x-1}=\frac{2}{0^+}=\infty.
$$
Part d
Repeatedly applying L’Hospital’s Rule, we have
\begin{align*}
\lim_{x\to 0}\frac{1-\cos 2x -2x^2}{x^4}
=&\ \lim_{x\to 0}\frac{2\sin 2x-4x}{4x^3} \\
=&\ \lim_{x\to 0}\frac{4\cos 2x-4}{12x^2} \\
=&\ \lim_{x\to 0}\frac{-8\sin 2x}{24x}\\
=&\ \lim_{x\to 0}\frac{-16\cos 2x}{24}\\
=&\ \frac{-16}{24}=-\frac{2}{3}.
\end{align*}
Another solution:
We can also do it using Exercise 30.2 (a). Namely,
$$
\lim_{u\to 0}\frac{u^3}{\sin u-u}=-6\quad \text{ or }\quad \lim_{u\to 0}\frac{\sin u-u}{u^3}=-\frac{1}{6}.
$$
Apply L’Hospital’s Rule and set $u=2x$. We have
\begin{align*}
\lim_{x\to 0}\frac{1-\cos 2x -2x^2}{x^4}
=&\ \lim_{x\to 0}\frac{2\sin 2x-4x}{4x^3} \\
=&\ \lim_{u\to 0}\frac{2\sin u-2 u}{\frac{1}{2}u^3} \\
=&\ 4\lim_{u\to 0}\frac{\sin u-u}{u^3}\\
=&\ -\frac{4}{6}=-\frac{2}{3}.
\end{align*}