Limit of the sine and logarithm function

Question

How do I calculate the following?

$$ \lim_{x \rightarrow 0^{+}} (-\ln(x) \sin(x)) $$

Answer 1

$$ \lim_{x \rightarrow 0^{+}} (-\ln(x) \sin(x)) = \lim_{x \rightarrow 0^{+}} \left(-x\ln(x) \cdot \frac{\sin x} x \right) $$ This is equal to $$ \left( \lim_{x \rightarrow 0^{+}} -x\ln(x) \right) \cdot \left( \lim_{x\to0^+} \frac{\sin x} x \right) $$ provided those both exist.

The first of those can be found by writing $\displaystyle\lim_{x\to0^+}\frac{-\ln x}{1/x}$ and applying L'Hopital's rule.

Answer 2

Do you recognize this limit as having a certain indeterminate form? Identify this form, and apply the standard technique from your textbook in order to deal with it. In case you don't have a textbook, this is $0\cdot \infty$ type. Rearrange the expression $\ln(x)\sin(x)$ to be $\frac{0}{0}$ form or $\frac{\infty}{\infty}$ form. Then use L'Hospital's Rule.

Answer 3

Use equivalents:

$\sin x\sim_0 x$, hence $\;-\ln x\sin x\sim_0 -x\ln x$, and it is a basic limit with logarithms that $$\lim_{x\to 0^+}x\ln x=0.$$ Furthermore, as $\ln x <0$ if $0<x<1$, we know $\lim\limits_{x\to 0^+}(-x\ln x)=0^+$.