Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 48

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

$\displaystyle \lim_{x \to 0} (\csc x - \cot x) = \lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right) = \lim_{x \to 0} \left( \frac{1 - \cos x}{\sin x} \right) = \frac{1- \cos 0}{\sin 0} = \frac{1-1}{0} = \frac{0}{0}$ Indeterminate.

Thus by applying L'Hospital's Rule...

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