# Finding $\lim_{x\rightarrow 0^+}\frac{\sin(x\log(x))}{x\log(x)}$ without L'Hospital's rule

Can someone help wit this question: Does this limit exist?

$$\lim_{x\rightarrow 0^+}\frac{\sin(x\log(x))}{x\log(x)}$$

I'm not allowed to use L'Hospital's rule, or differentiate or anything like that. I think I have to figure it out by using the epsilon-delta defintion or inequalities.

## 2 Answers

If one knows that $$\lim_{u \to 0}\frac {\sin u}u=1 \tag1$$ and that $$\lim_{x \to 0^+}x\ln x=0$$ then one may apply $(1)$ with $u=x\ln x$ to get, as $x \to 0^+$,

$$\lim_{x \to 0^+}\frac {\sin (x\ln x)}{(x\ln x)}=1. \tag2$$

• @tonytouch You are welcome. – Olivier Oloa Feb 15 '17 at 12:41
• @tonytouch If you find my answer useful, please feel free to accept it. Thanks. – Olivier Oloa Feb 16 '17 at 20:03
• Sorry, I didn't know I could do that xd – tonytouch Feb 17 '17 at 15:10
• @tonytouch No problem, thanks. – Olivier Oloa Feb 17 '17 at 15:32

Setting $$x\log(x)=t$$ and we have for $x$ tends to $$0^+$$ $$t$$ tends to $$0^+$$ thus we have $$\lim_{t \to 0^+}\frac{\sin(t)}{t}=1$$