MHB Limit involving non-indeterminate form

[Bueno]

I've been studying this kind of limits today and most of them were solved by the technique I mentioned in my previous topic, except for the one you just showed me how to solve and this one:

$$\lim_{x->0} (1 - cos(x))^{1/x}$$

(It approaches 0 from the left, but I don't know how to write it here)

I've tried some techniques, without success.
If you could show me a way, I'd be glad.

Thank you again,

Bueno

 

[MarkFL]

While we do allow two questions to be posted in a topic, usually the two questions are given in the first post, and so I decided for clarity to move this question to its own topic for clarity of discussion. :D

To express one-sided limits in $\LaTeX$, append either ^{-} or ^{+} to the value at which the variable is approaching, e.g.

$$\lim_{x\to0^{-}}(1-\cos(x))^{\frac{1}{x}}$$

This has the form $$0^{-\infty}=\frac{1}{0^{\infty}}$$ and so what is the end result?

 

[ZardoZ1]

As Mark said the limit as x approaches zero from the left it is infinite. So you will have to check what happens as x approaches zero from the right.

$$\lim_{x\to 0^{+}}\left(1-\cos(x)\right)^{\frac{1}{x}}=\lim_{x\to0^{+}}e^{log\left(\left(1-\cos(x)\right)^{\frac{1}{x}}\right)}=\lim_{x\to0^{+}}e^{\frac{log(1-\cos(x))}{x}}$$

and $$\lim_{x\to0^{+}}\frac{\log(1-\cos(x))}{x}=-\infty$$ so $$\lim_{x\to0^{+}}(1-\cos(x))^{\frac{1}{x}}=e^{-\infty}=\frac{1}{\infty}=0$$.

 

[Petrus]

I've been studying this kind of limits today and most of them were solved by the technique I mentioned in my previous topic, except for the one you just showed me how to solve and this one:

$$\lim_{x->0} (1 - cos(x))^{1/x}$$

(It approaches 0 from the left, but I don't know how to write it here)

I've tried some techniques, without success.
If you could show me a way, I'd be glad.

Thank you again,

Bueno

Hello Bueno,
as MarkFL and ZardoZ have helped you there is 'another' method to solve this
if we subsitute $$T=\frac{1}{x}$$ notice that we need to remake the limit
$$T=\frac{1}{0^-}=-\infty$$, $$T=\frac{1}{0^+}=\infty$$
so you can rewrite your limit
$$\lim_{T->\infty}(1-\cos(\frac{1}{T}))^T$$
$$\lim_{T->-\infty}(1-\cos(\frac{1}{T}))^T$$
does this make it easy?Can you continue? I hope that you understand my explain!

PS. Thanks for posting good problem :)
Regards,
$$|\pi\rangle$$