Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 50

Determine the $\displaystyle \lim_{x \to 0} \left( \cot x - \frac{1}{x}\right)$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.



$
\begin{equation}
\begin{aligned}
\lim_{x \to 0} \left( \cot x - \frac{1}{x}\right) &= \lim_{x \to 0} \left( \frac{\cos x}{\sin x} - \frac{1}{x} \right)\\
\\
&= \lim_{x \to 0} \frac{x \cos x - \sin x}{x \sin x}\\
\\
&= \frac{0(\cos 0) - \sin 0}{0 (\sin 0)}\\
\\
&= \frac{0}{0}
\end{aligned}
\end{equation}
$



Thus, by applying L'Hospitals Rule...

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0} \frac{x \cos x - \sin x}{x \sin x} &= \lim_{x \to 0} \frac{[x - (-\sin x) + \cos x] - \cos x}{x (\cos x) + \sin x}\\
\\
&= \lim_{x \to 0} \left[ \frac{-x \sin x}{x \cos x + \sin x} \right]
\end{aligned}
\end{equation}
$


If we evaluate the limit, we still get indeterminate form, so we must apply L'Hospitals Rule once more.. Hence,

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0} \left[ \frac{-x \sin x}{x \cos x + \sin x} \right] &= \lim_{x \to 0} \frac{[- x \cos x - \sin x]}{[x (-\sin x) + \cos x] + \sin x}\\
\\
&= \frac{-0(\cos 0) - \sin(0)}{0(-\sin(0)) + \cos 0 + \sin 0}\\
\\
&= \frac{0}{1} = 0
\end{aligned}
\end{equation}
$