What can I do this cos term to remove the divide by 0? I was asked to help someone with this problem, and I don't really know the answer why. But I thought I'd still try.
$$\lim_{t \to 10} \frac{t^2 - 100}{t+1} \cos\left( \frac{1}{10-t} \right)+ 100$$
The problem lies with the cos term. What can I do with the cos term to remove divide by 0 ? 
I found the answer to be $100$ (Google), but I do not know what they did to the $\cos$ term. Is that even the answer ?
Thanks!

The function is bounded below by $\left|\frac{t^2-100}{t+1}\right|(-1)+100$ and bounded above by  $\left|\frac{t^2-100}{t+1}\right|(+1)+100$.  Each approaches 100, so by the squeeze theorem the original limit is 100.
