Question

Compute the following limit. (Hint: Use squeeze theorem)

Answer 1

\[\lim_{x \rightarrow 0}\left[x~\cos\left(1+\frac{1}{x^2}\right)\right]\]

Answer 2

What was the joke from 17:57 - 18:03 in this video
https://www.youtube.com/watch?v=Rhn5uobZAyQ

Answer 3

I have no idea how I would use squeeze theorem. All the examples had two other functions and there was a graph and yeah mhmmm .-.

Answer 4

Use the fact that \(|\cos x|\le 1\). You can establish that
\[\left|x\cos\left(1+\frac{1}{x^2}\right)\right|\le|x|\]and the result of the limit should follow.

Answer 5

Is this the squeeze theorem? :x

Shouldn't we have two other functions that will have the same limit as our function at x = 0

Answer 6

I think from what he said, your two functions would be x and -x.

Answer 7

I'll brb

Answer 8

sooooo

\[\lim_{x \rightarrow 0}-x\le\lim_{x \rightarrow 0}\left[x~\cos\left(1+\frac{1}{x^2}\right)\right] \le \lim_{x \rightarrow 0}x\]

\[0\le\lim_{x \rightarrow 0}\left[x~\cos\left(1+\frac{1}{x^2}\right)\right] \le 0\]

and so \[\lim_{x \rightarrow 0}\left[x~\cos\left(1+\frac{1}{x^2}\right)\right] = \boxed{\bf{0}}\]

Answer 9

@zepdrix

Answer 10

The proof is not good enough to me, :(
I need know how to argue the function xcos(1+1/x^2) continuous everywhere to apply squezze theorem. However, 1/x^2 is undefined when x =0. and then cos (1+1/x^2) is....???

Answer 11

I agree. I was given a hint, but I really don't know how or where that came from and if that's an applicable step for finding it.

I do have to show my work for these questions.

Answer 12

Maybe @satellite73 @jim_thompson5910 @mathstudent55 @UnkleRhaukus can provide some more insight.

Answer 13

the cosine part is stuck between minus one and one

Answer 14

that is the point of this question

Answer 15

i didn't even look
you have it here \[\lim_{x \rightarrow 0}-x\le\lim_{x \rightarrow 0}\left[x~\cos\left(1+\frac{1}{x^2}\right)\right] \le \lim_{x \rightarrow 0}x\]

Answer 16

Ah, I see now.
\[ -1 \le \cos \left(1 + \frac{1}{x^2}\right)\le 1\]
\[ -1x \le x~\cos \left(1 + \frac{1}{x^2}\right)\le 1x\]
\[ -x \le x~\cos \left(1 + \frac{1}{x^2}\right)\le x\]

And then \[\lim_{x \rightarrow 0}-x\le\lim_{x \rightarrow 0}\left[x~\cos\left(1+\frac{1}{x^2}\right)\right] \le \lim_{x \rightarrow 0}x\]

I get it now. (:

Thanks everyone! :)