Proof for this power limit without differentiation

[wolly]

I am stuck at the last step and I don't know how to go on.Can someone help me?What do I do from the last step?

 

[pka]

I am stuck at the last step and I don't know how to go on.Can someone help me?What do I do from the last step?View attachment 12111

Here is a suggestion. Use \(\displaystyle \exp(x)=e^x\) to simplify notation.
The try \(\displaystyle \exp \left( {\log \left( {\cos {{\left( {\frac{1}{x}} \right)}^{{x^2}}}} \right)} \right) = \exp \left( {{x^2}\log \left( {\cos \left( {\frac{1}{x}} \right)} \right)} \right)\)

 

[Jomo]

Personally, I would use the squeeze theorem.
If x is large enough (you find out how large), then 0<cos(1/x)< .5 (.5 can be replaced with any other positive number less than 1)

 

[pka]

Personally, I would use the squeeze theorem.
If x is large enough (you find out how large), then 0<cos(1/x)< .5 (.5 can be replaced with any other positive number less than 1)

Jomo, \(\displaystyle \mathop {\lim }\limits_{x \to \infty } \cos \left( {\frac{1}{x}} \right) = 1\).
Do you want to rethink: x is large enough (you find out how large), then 0<cos(1/x)< .5???

 

[Jomo]

Jomo, \(\displaystyle \mathop {\lim }\limits_{x \to \infty } \cos \left( {\frac{1}{x}} \right) = 1\).
Do you want to rethink: x is large enough (you find out how large), then 0<cos(1/x)< .5???

Yes, I do want to rethink that one. Thanks for pointing it out. Good job!