# Does $\frac{\sin(x\ln x)}{x\ln x}$ as $x$ approaches $0$ from the right side have a limit?

I am trying to determine if

$$\lim_{x\rightarrow0^{+}}\frac{\sin(x \cdot \ln(x))}{x\cdot \ln(x)}$$

has a limit ?

Since

$$\ln(x) \rightarrow -\infty\text{ as }x \rightarrow 0^{+}$$

I have tried to use L'hôpital's rule with no luck. I was later told not to use L'hôpital's rule but instead use the $\epsilon -\delta$ definition to get the result

$$\lim_{x\rightarrow 0^{+}} \frac{\sin(x \cdot \ln(x))}{x\cdot \ln(x)}=1$$

I would like some help in proving the limit of the function by using the $\epsilon -\delta$ definition

• Using $\epsilon, \delta$ here seems like a waste of time.
Let $u = x \ln x$. We have $$\lim_{x \to 0^+} u = 0,$$ and $$\lim_{u \to 0}\frac{\sin u}{u} = 1.$$ You can prove both limits with L'hôpital's rule. For the first one, apply the rule to $\ln x/(1/x)$. Therefore $$\lim_{x \to 0^+}\frac{\sin(x \ln x)}{x \ln x} = 1.$$