Limit involving non-indeterminate form

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In summary, in this equation, $\pi\rangle$ is the eleven's complement of $\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)+\left
  • #1
Bueno
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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
 
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  • #2
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?
 
  • #3
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 $$\lim_{x\to0^{+}}(1-\cos(x))^{\frac{1}{x}}=e^{-\infty}=\frac{1}{\infty}=0$$.
 
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  • #4
Bueno said:
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,
\(\displaystyle |\pi\rangle\)
 

FAQ: Limit involving non-indeterminate form

What is a limit involving non-indeterminate form?

A limit involving non-indeterminate form is a type of mathematical expression that cannot be evaluated directly because it results in an undefined or infinite value.

What are the different types of non-indeterminate forms?

The three main types of non-indeterminate forms are 0/0, ∞/∞, and ∞-∞.

How do you evaluate a limit involving non-indeterminate form?

To evaluate a limit involving non-indeterminate form, you must manipulate the expression algebraically to simplify it into a form that can be evaluated using basic limit laws.

Can all limits involving non-indeterminate form be evaluated?

No, not all limits involving non-indeterminate form can be evaluated. Some may result in an undefined value or may require more advanced techniques to evaluate.

What are some common methods for evaluating limits involving non-indeterminate form?

Some common methods for evaluating limits involving non-indeterminate form include factoring, using L'Hopital's rule, and using trigonometric identities.

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