I'm trying to compute
$$ \lim_{x\to 0^+} x^{\sin(x)}. $$
I've been trying to get it into a form where I can apply L'Hopital's rule, but I haven't had any success. Namely, I haven't been able to massage it into a fraction form. Any advice?
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The function f(x)=e^sin(x) has a Taylor series centered at zero but this x^sin(x) looks much more difficult. I hope I'm wrong! – user138027 Apr 24 '14 at 00:15
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Alternatively write the taylor polynomial of this at $x = 1$. Then the conclusion also follows. – Sandeep Silwal Apr 23 '14 at 23:47
4 Answers
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Hint Using the L'Hopital's rule:
$$\lim_{x\to0}\sin x\ln x=\lim_{x\to0}\frac{\ln x}{\frac1{\sin x}}=-\lim_{x\to0}\frac{\sin x}{x}\frac{\sin x}{\cos x}=0$$
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We have $\lim_{x \to 0^+} \sin(x) \log(x) = 0$. Hence, $$\lim_{x \to 0^+} x^{\sin(x)} = 1$$
user141421
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1Is the first fact supposed to be evident, or is it an implied use of L'Hopital's to find the limit? – MT_ Apr 24 '14 at 00:06
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@MichaelT Yes, it is evident since as $x \to 0$, $\sin(x) \sim x$ and $\lim_{x \to 0^+} x \log(x) = 0$. – user141421 Apr 24 '14 at 02:49
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Write $$x^{\sin x}=e^{\ln x\sin x}$$ Then, apply l'Hopital's rule to $$\frac{\sin x}{1/\ln x}$$ Last, apply the continuity of exponential function.
ajotatxe
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$\mathbf{Result \; I}:$ If $x \in [0,\pi)$, then
$$ x^{\sin x} \leq \sin x(x - 1 ) + 1 $$
To see why this is true, notice that $x \in [0, \pi) $ implies that $\sin x \geq 0 $, hence invoking the Young's Inequality and the fact that $\sin x + (1 - \sin x) = 1$, we have
$$ x^{\sin x} = x^{\sin x}\cdot1^{1- \sin x} \leq \sin x \cdot x + 1 - \sin x = \sin x(x -1 ) + 1 $$
$\mathbf{Result \; II}$: $$x^{\sin x} \geq 1 - 10 x $$
You should try to show this inequality: You can find the min of $f(x) = x^{ \sin x} + 10 x $.
Now,
$$ \sin x(x -1) + 1 \geq x^{\sin x} \geq 1 - 10 x $$
Hence, an application of of the squeeze rule gives that
$$ \lim_{x \to 0^+} x^{\sin x} = 1 $$