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How can I find that specific limit?

I've tried the following:

$f(x) = |x|^{\sin {x}} = e^{\sin x \cdot \ln(|x|)} = e^{\frac{\ln |x|}{\frac{1}{x}}}$

And use LLhopittal but I don't think it's correct since $\lim_{x\to0}\ln(|x|)=-\infty$ and $\lim_{x\to0}\frac{1}{x}=\infty$

4 Answers4

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You do not need L'Hopital. Just rewrite $$ f(x) = |x|^{\sin {x}} = e^{\sin x \cdot \ln(|x|)} =e^{\frac{\sin x}{x} \cdot x\ln(|x|)}\ , $$ and use the standard limits $$\sin(x)/x\to 1$$ and $$x\log |x|\to 0$$ as $x\to 0$, to conclude that your function tends to $e^0=1$.

Pierpaolo Vivo
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  • I'm not familiar of the limit of $x\log|x|$. Could you show it's calculation? –  Jan 29 '17 at 19:34
  • You just need to compute the two limits $\lim_{x\to 0^+} x\log x$ and $\lim_{x\to 0^-}x \log (-x)$ and show that they are equal. For each of these, you can use http://math.stackexchange.com/questions/522973/lim-x-to0-x-ln-x-without-lhopitals-rule [if you don't want to use L'Hopital - which as I said is not needed, and my personal recommendation would be to try to avoid it whenever possible, at least in the 'learning' phase, as in my opinion it encourages 'lazyness'] or http://math.stackexchange.com/questions/470952/limit-of-x-log-x-as-x-tends-to-0 – Pierpaolo Vivo Jan 29 '17 at 20:37
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The easiest way to get rid of the absolute value is using lateral limits: $$\lim_{x\to 0^+} |x|^{\sin x}= \lim_{x\to 0^+} x^{\sin x}$$ now you can apply l'Hôpital normally.

A. Salguero-Alarcón
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Appears the indeterminate form $0^0.$ Using $\sin x\sim x$ as $x\to 0:$ $$\begin{aligned} \lambda=\lim_{x\to 0^+}\log |x|^{\sin x}=\lim_{x\to 0^+}\sin x\log x=\lim_{x\to 0^+}x\log x\\ =\lim_{x\to 0^+}\frac{\log x}{1/x}=\left\{\frac{-\infty}{+\infty}\right\}=\lim_{x\to 0^+}\frac{1/x}{-1/x^2}=\lim_{x\to 0^+}\;(-x)=0. \end{aligned}$$ $$\Rightarrow \lim_{x\to 0^+}|x|^{\sin x}=e^\lambda=1$$ On the other hand

$$\begin{aligned} \mu=\lim_{x\to 0^-}\log |x|^{\sin x}=\lim_{x\to 0^-}\sin x\log (-x)=\lim_{x\to 0^-}x\log (-x)\\ =\lim_{x\to 0^-}\frac{\log (-x)}{1/x}=\left\{\frac{-\infty}{+\infty}\right\}=\lim_{x\to 0^-}\frac{1/x}{-1/x^2}=\lim_{x\to 0^-}\;(-x)=0. \end{aligned}$$ $$\Rightarrow \lim_{x\to 0^-}|x|^{\sin x}=e^\mu=1$$ So, $\displaystyle\lim_{x\to0}|x|^{\sin x}=1.$

Fernando Revilla
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$\lim_{x\to0}|x|^{\sin x}$

Getting rid of absolute value gives us

$\lim_{x\to0+}(x^{\sin x})$

Remember this rule? $$a^x=e^{\ln \left(a^x\right)}=e^{x\cdot \ln \left(a\right)}$$

$$\lim _{x\to \:0}\left(e^{\sin \left(x\right)\ln \left(x\right)}\right)$$

This is a better form I thik, that you can apply l'hoptial

K Split X
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