# Squeeze/Sandwhich theorem

1. Oct 4, 2007

### skateza

1. The problem statement, all variables and given/known data
Use the Squeeze/Sandwich theorem to show the

$$limx->1 [(x-1)^{2}sin(\frac{1}{1-x})]=0$$

Thats the lim as x approaches 1 for the whole square bracket

3. The attempt at a solution
i split it up into the two seperate limits, but i'm kinda lost on how the sandwhich theorem works, my professor said something along the lines of sin1/1-x has to be between 1 and -1 because all sin are, than he change that to an x and found the limit as x approached zero and the two limits worked out, but this limit has x approaching 1 so i'm stuck.

2. Oct 4, 2007

### Dick

If we let L be the form you are taking the limit of, then -(x-1)^2<=L<=+(x-1)^2, since the sin is always between -1 and +1, right? What are the outside limits?

3. Oct 4, 2007

### skateza

ok i solved it, it's 0... but i kinda just put what the data should be to obtain zero, how do u find the limit as x approaces 1 for sin(1/(1-x)), how can u tell what the sin of infinity is?

4. Oct 4, 2007

### Dick

It oscillates between -1 and 1. It doesn't have a limit. L oscillates too, but the amplitude of the oscillations get smaller and smaller as it approaches infinity.