# Thread: sandwhich/squeeze theorm..help pls =]

1. ## sandwhich/squeeze theorm..help pls =]

hi, im unsure on how to solve a certain limit using the sandwhich rule.

lim [(x-3) sin(1/(3-x))] as x -> 3

any help or hints would be much appreciated..thank you

2. Originally Posted by tasukete
hi, im unsure on how to solve a certain limit using the sandwhich rule.

lim [(x-3) sin(1/(3-x))] as x -> 3

any help or hints would be much appreciated..thank you
The limit can be re-written as

$\lim_{y \rightarrow 0} (-y) \, \sin\left( \frac{1}{y} \right) = -\lim_{y \rightarrow 0}y \, \sin\left( \frac{1}{y} \right)$

where y = 3 - x.

This in turn can be re-writen as $- \lim_{t \rightarrow \infty} \frac{\sin t}{t}$

where $t = \frac{1}{y}$.

3. just need to confirm something....if y = x - 3

wouldnt that make -y = 3-x ??? since u put:

ysin(1/y) wouldnt it be: ysin(1/-y) instead? hope im right..

4. oops :O pls ignore my previous post...read it in such a rush didnt see the -ve sign ...sry =] . thank you for your help so far