Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 42

Determine the $\displaystyle \lim_{x \to 0^+} \sin x \ln x$. 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^+} \sin x \ln x = \lim_{x \to 0^+} \frac{\ln x}{\csc x} $


By applying L'Hospital's Rule...

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0^+} \frac{\ln x}{\csc x} &= \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\csc x \cot x} \\
\\
&= \lim_{x \to 0^+} \frac{\left( \frac{1}{x} \right)}{-\frac{1}{\sin x} \left( \frac{\cos x}{\sin x} \right)}\\
\\
&= \lim_{x \to 0^+} \frac{- \sin^2 x}{x \cos x}
\end{aligned}
\end{equation}
$


We can rewrite the limit as...

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0^+} \frac{-\sin^2x}{x \cos x} &= - \lim_{x \to 0^+} \left( \frac{\sin x}{x} \cdot \frac{\sin x}{\cos x} \right)\\
\\
&= - \left[ \lim_{x \to 0^+} \frac{\sin x}{x} \cdot \lim_{x \to 0^+} \tan x \right]\\
\\
\text{Recall that } \lim_{x \to 0^+} \frac{\sin x}{x} &= 1 \text{ ,so...}\\
\\
&= - \left[1 \cdot \lim_{x \to 0^+} \tan x \right]\\
\\
&= -[1 \cdot 0]\\
\\
&= 0
\end{aligned}
\end{equation}
$