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Limit involving non-indeterminate form

Bueno

New member
Mar 29, 2013
14
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:

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

(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
 
Last edited by a moderator:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.

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

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

ZardoZ

New member
Apr 15, 2013
1
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 $$\lim_{x\to0^{+}}(1-\cos(x))^{\frac{1}{x}}=e^{-\infty}=\frac{1}{\infty}=0$$.
 
Last edited:

Petrus

Well-known member
Feb 21, 2013
739
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:

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

(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 \(\displaystyle T=\frac{1}{x}\) notice that we need to remake the limit
\(\displaystyle T=\frac{1}{0^-}=-\infty\), \(\displaystyle T=\frac{1}{0^+}=\infty\)
so you can rewrite your limit
\(\displaystyle \lim_{T->\infty}(1-\cos(\frac{1}{T}))^T\)
\(\displaystyle \lim_{T->-\infty}(1-\cos(\frac{1}{T}))^T\)
does this make it easy?Can you continue? I hope that you understand my explain!

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