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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 2

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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 $x \to 0^+$,

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

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    @tonytouch You are welcome. – 2017-02-15
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    @tonytouch If you find my answer useful, please feel free to accept it. Thanks. – 2017-02-16
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    Sorry, I didn't know I could do that xd – 2017-02-17
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    @tonytouch No problem, thanks. – 2017-02-17
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Setting $$x\log(x)=t$$ and we have for $x$ tends to $$0^+$$ $$t$$ tends to $$0^+$$ thus we have $$\lim_{t \to 0^+}\frac{\sin(t)}{t}=1$$