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.


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 $x \to 0^+$,

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

  • $\begingroup$ @tonytouch You are welcome. $\endgroup$ – Olivier Oloa Feb 15 '17 at 12:41
  • $\begingroup$ @tonytouch If you find my answer useful, please feel free to accept it. Thanks. $\endgroup$ – Olivier Oloa Feb 16 '17 at 20:03
  • $\begingroup$ Sorry, I didn't know I could do that xd $\endgroup$ – tonytouch Feb 17 '17 at 15:10
  • $\begingroup$ @tonytouch No problem, thanks. $\endgroup$ – 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 $$\lim_{t \to 0^+}\frac{\sin(t)}{t}=1$$


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