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amoodarya
 one year ago
The way to answer this limit
amoodarya
 one year ago
The way to answer this limit

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amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\to 0} \sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)}  \sin(x)\log(\sin(x))\right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0try direct substitution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think every thing will be 0

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0I can do that in very long terms ... but I looking for a good idea !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok but it is not undefined by direct substitution but 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.2maybe we can use this... \[u=\sin(x)^{\cos(x)} \\ \frac{u'}{u}=\frac{\cos^2(x)}{\sin(x)}\sin(x) \ln(\sin(x))\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0then 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.2i was thinking about trying to somehow use l'hosptial backwards if you know what I mean

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow 0} \frac{u'}{1}=\lim_{x \rightarrow 0} \frac{u}{x}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2we need to show this \[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)}=1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2oops u does goes to 0

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim sinx^{\cos x} \rightarrow 0 (x \to 0)\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ \begin{align*} L&=\lim_{x\to 0} \sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)}  \sin(x)\log(\sin(x))\right)\\&=\lim_{x\to 0} \cos^2(x)\sin(x)^{\cos(x)1}  \lim_{x\to 0} \sin(x)^{\cos(x)+1}\log(\sin(x))\\ \end{align*} \] By Mathematica, both limits exists. No idea how to evaluate though.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1The first one is 1 and the second one is 0.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1In fact, \[ \lim_{x\to 0}\sin(x)^{\cos(x)1}=1 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)}\] hmmm near x we have sin(x) is approximately x I don't know if we can do this...\[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)} =\lim_{x \rightarrow 0}\frac{1}{x} x ^{\cos(x)} =\lim_{x \rightarrow 0}x^{\cos(x)1}\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Squeeze theorem using x and x?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[v=x^{\cos(x)1} \\ \ln(v)=(\cos(x)1) \ln(x) \\ \lim_{x \rightarrow 0} (\cos(x)1)\ln(x)=\lim_{x \rightarrow 0}\frac{\ln(x)}{\frac{1}{\cos(x)1}} \\ =\lim_{x \rightarrow 0}\frac{\frac{1}{x}}{\frac{\sin(x)}{(\cos(x)1)^2}} =\lim_{x \rightarrow 0}\frac{1(\cos(x)1)^2}{x \sin(x)} \\ = \lim_{x \rightarrow 0} \frac{1 (\frac{\cos(x)1}{x})^2}{\frac{x}{x} \frac{\sin(x)}{x}}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[=\frac{1(0)^2}{1(1)}=0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[e^{\ln(v)}>e^{0}=1\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.0thank U all . I get it

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1The problem is that \[ \lim_{x\to 0}\sin(x)^{\cos(x)1} \quad\text{ does not exist on }\mathbb{R}\\ \lim_{x\to 0^+}\sin(x)^{\cos(x)1}=1 \]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I don't think the left hand limit exists on real number.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2thomas 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 \[\sin(x)\approx x \text{ for } x \approx 0 \\ \cos(x) \approx 1 \text{ for } x \approx 0 \\ \sin(x)^{\cos(x)}(\frac{\cos^2(x)}{\sin(x)}\sin(x)\ln(\sin(x))) \approx x^{1}(\frac{1^2}{x}x \ln(x)) \\ \lim_{x \rightarrow 0^+} x(\frac{1}{x}x \ln(x)) \\ \\ =\lim_{x \rightarrow 0^+}(1x^2 \ln(x))\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2like we can use l'hospital on that second term to show it goes to 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.2then we have 10 which is 1

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I mean \[ \sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)}  \sin(x)\log(\sin(x))\right) \; \text{does not exist for } x\leq 0 \]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1So there is no left hand limit and no limit. At least not on real numbers.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but we can we find the right hand limit

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I changed those little thingys above to + signs to denote I was looking at the right hand limit
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