The way to answer this limit

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The way to answer this limit

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\[\lim_{x\to 0} \sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)} - \sin(x)\log(\sin(x))\right)\]
try direct substitution
i think every thing will be 0

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the answer =0
answer is +1
I can do that in very long terms ... but I looking for a good idea !
ok but it is not undefined by direct substitution but 0
maybe we can use this... \[u=\sin(x)^{\cos(x)} \\ \frac{u'}{u}=\frac{\cos^2(x)}{\sin(x)}-\sin(x) \ln(\sin(x))\]
then we simply need to find \(\lim\limits_{x\to 0}~~u' \) is it http://www.wolframalpha.com/input/?i=lim+%28x%5Cto+0%29+%28%28sin%28x%29%29%5E%28cos%28x%29%29%29%27
i was thinking about trying to somehow use l'hosptial backwards if you know what I mean
\[\lim_{x \rightarrow 0} \frac{u'}{1}=\lim_{x \rightarrow 0} \frac{u}{x}\]
we need to show this \[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)}=1\]
oops u does goes to 0
\[\lim sinx^{\cos x} \rightarrow 0 (x \to 0)\]
\[ \begin{align*} L&=\lim_{x\to 0} \sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)} - \sin(x)\log(\sin(x))\right)\\&=\lim_{x\to 0} \cos^2(x)\sin(x)^{\cos(x)-1} - \lim_{x\to 0} \sin(x)^{\cos(x)+1}\log(\sin(x))\\ \end{align*} \] By Mathematica, both limits exists. No idea how to evaluate though.
The first one is 1 and the second one is 0.
In fact, \[ \lim_{x\to 0}\sin(x)^{\cos(x)-1}=1 \]
\[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)}\] hmmm near x we have sin(x) is approximately x I don't know if we can do this...\[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)} =\lim_{x \rightarrow 0}\frac{1}{x} x ^{\cos(x)} =\lim_{x \rightarrow 0}x^{\cos(x)-1}\]
near x=0*
Squeeze theorem using x and -x?
\[v=x^{\cos(x)-1} \\ \ln(v)=(\cos(x)-1) \ln(x) \\ \lim_{x \rightarrow 0} (\cos(x)-1)\ln(x)=\lim_{x \rightarrow 0}\frac{\ln(x)}{\frac{1}{\cos(x)-1}} \\ =\lim_{x \rightarrow 0}\frac{\frac{1}{x}}{\frac{-\sin(x)}{(\cos(x)-1)^2}} =\lim_{x \rightarrow 0}\frac{-1(\cos(x)-1)^2}{x \sin(x)} \\ = \lim_{x \rightarrow 0} \frac{-1 (\frac{\cos(x)-1}{x})^2}{\frac{x}{x} \frac{\sin(x)}{x}}\]
\[=\frac{-1(0)^2}{1(1)}=0\]
\[e^{\ln(v)}->e^{0}=1\]
so v->1
thank U all . I get it
The problem is that \[ \lim_{x\to 0}\sin(x)^{\cos(x)-1} \quad\text{ does not exist on }\mathbb{R}\\ \lim_{x\to 0^+}\sin(x)^{\cos(x)-1}=1 \]
I don't think the left hand limit exists on real number.
thomas I see what you mean we need x>0 for ln(x) to exist also I think this approach might be better than my earlier if it is allowed \[\sin(x)\approx x \text{ for } x \approx 0 \\ \cos(x) \approx 1 \text{ for } x \approx 0 \\ \sin(x)^{\cos(x)}(\frac{\cos^2(x)}{\sin(x)}-\sin(x)\ln(\sin(x))) \approx x^{1}(\frac{1^2}{x}-x \ln(x)) \\ \lim_{x \rightarrow 0^+} x(\frac{1}{x}-x \ln(x)) \\ \\ =\lim_{x \rightarrow 0^+}(1-x^2 \ln(x))\]
like we can use l'hospital on that second term to show it goes to 0
then we have 1-0 which is 1
I mean \[ \sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)} - \sin(x)\log(\sin(x))\right) \; \text{does not exist for } x\leq 0 \]
right
So there is no left hand limit and no limit. At least not on real numbers.
but we can we find the right hand limit
I changed those little thingys above to + signs to denote I was looking at the right hand limit

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