Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 20

Determine the $\displaystyle \lim_{x \to 1} \frac{\ln x}{\sin \pi x}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

$\displaystyle \lim_{x \to 1} \frac{\ln x}{\sin \pi x} = \frac{\ln(1)}{\sin[(\pi)(1)]} = \frac{0}{\sin \pi} = \frac{0}{0} \text{ Indeterminate form}$

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

$
\begin{equation}
\begin{aligned}
\lim_{x \to 1} \frac{\ln x}{\sin \pi x} &= \lim_{x \to 1} \frac{\frac x{d}{dx}(\ln x)}{\frac{d}{dx}\sin \pi (x)}\\
\\
&= \lim_{x \to 1} \frac{\frac{1}{x}}{\cos \pi x \cdot \pi}\\
\\
&= \lim_{x \to 1} \frac{1}{\pi x \cos (\pi x)}\\
\\
&= \frac{1}{\pi (1) \cos (\pi(1))}\\
\\
&= \frac{1}{\pi \cos (\pi)}\\
\\
&= \frac{1}{\pi(-1)}\\
\\
&= \frac{-1}{\pi}
\end{aligned}
\end{equation}
$