Question

The way to answer this limit

Answer 1

\[\lim_{x\to 0} \sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)} - \sin(x)\log(\sin(x))\right)\]

Answer 2

try direct substitution

Answer 3

i think every thing will be 0

Answer 4

the answer =0

Answer 5

answer is +1

Answer 6

I can do that in very long terms ... but I looking for a good idea !

Answer 7

ok but it is not undefined by direct substitution but 0

Answer 8

maybe we can use this...
\[u=\sin(x)^{\cos(x)} \\ \frac{u'}{u}=\frac{\cos^2(x)}{\sin(x)}-\sin(x) \ln(\sin(x))\]

Answer 9

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

Answer 10

i was thinking about trying to somehow use l'hosptial backwards if you know what I mean

Answer 11

\[\lim_{x \rightarrow 0} \frac{u'}{1}=\lim_{x \rightarrow 0} \frac{u}{x}\]

Answer 12

we need to show this
\[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)}=1\]

Answer 13

oops u does goes to 0

Answer 14

\[\lim sinx^{\cos x} \rightarrow 0 (x \to 0)\]

Answer 15

\[
\begin{align*}
L&=\lim_{x\to 0} \sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)} - \sin(x)\log(\sin(x))\right)\\&=\lim_{x\to 0} \cos^2(x)\sin(x)^{\cos(x)-1} - \lim_{x\to 0} \sin(x)^{\cos(x)+1}\log(\sin(x))\\
\end{align*}
\]
By Mathematica, both limits exists. No idea how to evaluate though.

Answer 16

The first one is 1 and the second one is 0.

Answer 17

In fact,
\[
\lim_{x\to 0}\sin(x)^{\cos(x)-1}=1
\]

Answer 18

\[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)}\]
hmmm near x we have sin(x) is approximately x
I don't know if we can do this...\[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)} =\lim_{x \rightarrow 0}\frac{1}{x} x ^{\cos(x)} =\lim_{x \rightarrow 0}x^{\cos(x)-1}\]

Answer 19

near x=0*

Answer 20

Squeeze theorem using x and -x?

Answer 21

\[v=x^{\cos(x)-1} \\ \ln(v)=(\cos(x)-1) \ln(x) \\ \lim_{x \rightarrow 0} (\cos(x)-1)\ln(x)=\lim_{x \rightarrow 0}\frac{\ln(x)}{\frac{1}{\cos(x)-1}} \\ =\lim_{x \rightarrow 0}\frac{\frac{1}{x}}{\frac{-\sin(x)}{(\cos(x)-1)^2}} =\lim_{x \rightarrow 0}\frac{-1(\cos(x)-1)^2}{x \sin(x)} \\ = \lim_{x \rightarrow 0} \frac{-1 (\frac{\cos(x)-1}{x})^2}{\frac{x}{x} \frac{\sin(x)}{x}}\]

Answer 22

\[=\frac{-1(0)^2}{1(1)}=0\]

Answer 23

\[e^{\ln(v)}->e^{0}=1\]

Answer 24

so v->1

Answer 25

thank U all . I get it

Answer 26

The problem is that \[
\lim_{x\to 0}\sin(x)^{\cos(x)-1} \quad\text{ does not exist on }\mathbb{R}\\
\lim_{x\to 0^+}\sin(x)^{\cos(x)-1}=1
\]

Answer 27

I don't think the left hand limit exists on real number.

Answer 28

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
\[\sin(x)\approx x \text{ for } x \approx 0 \\ \cos(x) \approx 1 \text{ for } x \approx 0 \\ \sin(x)^{\cos(x)}(\frac{\cos^2(x)}{\sin(x)}-\sin(x)\ln(\sin(x))) \approx x^{1}(\frac{1^2}{x}-x \ln(x)) \\ \lim_{x \rightarrow 0^+} x(\frac{1}{x}-x \ln(x)) \\ \\ =\lim_{x \rightarrow 0^+}(1-x^2 \ln(x))\]

Answer 29

like we can use l'hospital on that second term to show it goes to 0

Answer 30

then we have 1-0 which is 1

Answer 31

I mean \[
\sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)} - \sin(x)\log(\sin(x))\right) \; \text{does not exist for } x\leq 0
\]

Answer 32

right

Answer 33

So there is no left hand limit and no limit. At least not on real numbers.

Answer 34

but we can we find the right hand limit

Answer 35

I changed those little thingys above to + signs to denote I was looking at the right hand limit