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

1. amoodarya

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

2. anonymous

try direct substitution

3. anonymous

i think every thing will be 0

4. anonymous

5. amoodarya

6. amoodarya

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

7. anonymous

ok but it is not undefined by direct substitution but 0

8. freckles

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

9. ganeshie8

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

10. freckles

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

11. freckles

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

12. freckles

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

13. freckles

oops u does goes to 0

14. amoodarya

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

15. thomas5267

\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.

16. thomas5267

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

17. thomas5267

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

18. freckles

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

19. freckles

near x=0*

20. thomas5267

Squeeze theorem using x and -x?

21. freckles

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

22. freckles

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

23. freckles

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

24. freckles

so v->1

25. amoodarya

thank U all . I get it

26. thomas5267

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

27. thomas5267

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

28. freckles

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

29. freckles

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

30. freckles

then we have 1-0 which is 1

31. thomas5267

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

32. freckles

right

33. thomas5267

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

34. freckles

but we can we find the right hand limit

35. freckles

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