# 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{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}

We can rewrite the limit as...

\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}

Posted on