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Solution:

Part a

We can apply L’Hospital’s Rule repeatedly to get the answer. Please check the conditions for applying L’Hospital’s Rule carefully for every step. We have
\begin{align*}
\lim_{x\to 0}\frac{x^3}{\sin x-x}
=&\ \lim_{x\to 0}\frac{3x^2}{\cos x-1}\\
=&\ \lim_{x\to 0}\frac{6x}{-\sin x}\\
=&\ \lim_{x\to 0}\frac{6}{-\cos x}\\
=&\ \frac{6}{-1}=-6.
\end{align*}

Part b

We shall use the following limit from Example 1, that is
$$
\lim_{x\to 0}\frac{\sin x}{x}=1.
$$ This also implies that
\begin{equation}\label{eq:30-2}
\lim_{x\to 0}\frac{\tan x}{x}=\lim_{x\to 0}\frac{\sin x}{x}\lim_{x\to 0}\frac{1}{\cos x}=1.
\end{equation} Therefore, we have
\begin{align*}
\lim_{x\to 0}\frac{\tan x-x}{x^3}
=&\ \lim_{x\to 0}\frac{\sec^2 x-1}{3x^2}\\
=&\ \lim_{x\to 0}\frac{\tan^2 x}{3x^2}\\
=&\ \frac{1}{3}\lim_{x\to 0}\Big(\frac{\tan x}{x}\Big)^2\\
=&\ \frac{1}{3}\Big(\lim_{x\to 0}\frac{\tan x}{x}\Big)^2\\
\text{ Use \eqref{eq:30-2} }=&\ \frac{1}{3}\cdot 1=\frac{1}{3}.
\end{align*}

Part c

We can apply L’Hospital’s Rule repeatedly to get the answer. Please check the conditions for applying L’Hospital’s Rule carefully for every step. We have
\begin{align*}
\lim_{x\to 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right)
=&\ \lim_{x\to 0}\frac{x-\sin x}{x\sin x}\\
=&\ \lim_{x\to 0}\frac{1-\cos x}{\sin x+x\cos x}\\
=&\ \lim_{x\to 0}\frac{\sin x}{\cos x+\cos x-x\sin x}\\
=&\ \frac{0}{1+1-0}=0.
\end{align*}

Part d

We have
$$
\ln(\cos x)^{1/x^2}=\frac{1}{x^2}\ln(\cos x).
$$ Instead of computing $\lim_{x\to 0} (\cos x)^{1/x^2}$ directly, we compute $\lim_{x\to 0}\dfrac{1}{x^2}\ln(\cos x)$ first. We can apply L’Hospital’s Rule and obtain that
\begin{align*}
\lim_{x\to 0}\frac{1}{x^2}\ln(\cos x)
=&\ \lim_{x\to 0}\frac{\ln(\cos x)}{x^2}\\
=&\ \lim_{x\to 0}\frac{ \frac{-\sin x}{\cos x} }{2x}\\
=&\ -\frac{1}{2}\lim_{x\to 0}\frac{\tan x}{x}\\
\text{ Use \eqref{eq:30-2} }=&\ -\frac{1}{2}.
\end{align*} Therefore, by Theorem 20.5, we have
$$
\lim_{x\to 0}\ln(\cos x)^{1/x^2}=e^{-1/2}=\frac{1}{\sqrt e}.
$$