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 cos term is irrelevant. It can only wiggle between $-1$ and $1$, and is therefore killed by the $t^2-100$ term, since that approaches $0$.
For a less cluttered version of the same phenomenon, consider the function $f(x)=x\sin\left(\frac{1}{x}\right)$ (for $x\ne 0$). The absolute value of this is always $\le |x|$, so (by Squeezing) $f(x)\to 0$ as $x\to 0$.
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.
