$\lim_{x \to 0} \sin x ^{\sin x} $ Hi, Help me do it please. I am asking for any advices, helpful observations. Thanks in advance.
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1First of all, note that the expresseion makes little sense for negative (but almost $0$) values of $x$ – Hagen von Eitzen Jan 27 '15 at 17:08
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1look at $\lim_{x \to 0}x^x$ first. – abel Jan 27 '15 at 17:16
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$$\sin (x)^{\sin (x)}=e^{\sin (x)\ln(\sin(x))}$$
Moreover $\sin x\sim x$ if $x\to 0$, therefore $$e^{\sin(x)\ln(\sin x)}\sim e^{x\ln x}.$$
But $\lim_{x\to 0^+}x\ln x=0$ and thus
$$ \lim_{x\to 0^+}e^{x\ln x}=e^0=1.$$
Notice: your function is not define on $0^-$.
idm
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$$x\ln x=\frac{\ln x}{\frac{1}{x}}.$$ By l'hopital, $$\lim_{x\to 0^+}x\ln x=\lim_{x\to 0^+}\frac{\ln x}{\frac{1}{x}}=\lim_{x\to 0^+}(-x)=0.$$ But there is many way to prove it... – idm Jan 27 '15 at 18:56
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We have $\sin x\to 0$ as $x\to 0$, hence you may consider $\lim_{x\to0^+}x^x$ instead.
Hagen von Eitzen
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$y=\sin x^{\sin x}$
$\log y=\sin x \log{\sin x}$
$$\log y={\frac{\log {\sin x}}{\frac{1}{\sin x}}}$$
Now you can use l-hopital rule as as $x \to 0$, $log \sin x \to -\infty $ also $\frac{1}{\sin x} \to \infty$
Can you do it now?
Shobhit
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