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## Compute limits using L’Hospital’s Rule III

Solution:

### Part a

Since $|\sin x|\leq 1$, we have
\label{30-3-1}
\lim_{x\to \infty}\frac{\sin x}{x}=0.
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*}