Question

find the following limit........lim as x approaches 1 (x-1)^2cos(1/(x-1))

Answer 1

Use squeeze theorem to show that the limit goes to zero.

Answer 2

I'm still having trouble it's the cos part that throws me off

Answer 3

Don't worry about the cosine. Since \[\left| \cos (\frac {1}{1-x}) \right| \le 1\]Then\[\left| (x -1)^2 \cos (\frac {1}{1-x}) \right| = \left| (x -1)^2 \right| \times \left| \cos (\frac {1}{1-x}) \right| \le \left| (x -1)^2 \right| \times 1 = (x -1)^2 \]And thus,\[- (x -1)^2 \le (x -1)^2\cos (\frac {1}{1-x}) \le (x -1)^2\]Because \[\lim_{x \rightarrow 1} - (x -1)^2 = 0 = \lim_{x \rightarrow 1} (x -1)^2\]We can say the limit as x--1 of (x-1)^2cos(1/(x-1)) is squeezed in between the 2 known limits as is essentially the same limit.
So by squeeze/sandwich theorem,\[\lim_{x \rightarrow 1} (x -1)^2\cos (\frac {1}{1-x}) = 0\]

Answer 4

then the limit doesn't exist!!! am I right?

Answer 5

The functional value doesn't exist, but the limit does exist.

Answer 6

ohhh okay.....the limit is zero?

Answer 7

Yup. Here's a visual if you wanna see what's happening/what the squeeze theorem is about.
http://www.wolframalpha.com/input/?i=plot+-%28x-1%29^2%2C+%28x-1%29^2+cos%281%2F%28x-1%29%29%2C+%28x-1%29^2+from+.85+to+1.15

Answer 8

I apreciate all your help you are the best!!!!

Answer 9

Glad I could help =)