Complicated squeeze theorem This was a question I had on one of my exams and I just could not figure out how to do it, any explanation would be greatly appreciated!
Let $f$ be a function such that $-3 \le f(x) \le 5.$ Find $\lim_{x\to2}((x-2)^2f(x)+1)$  
 A: The answer given above is correct, of course.  I hope to add a bit of insight to help you analyze these kinds of questions.  Also, depending on the level of your course, you might be epected to provide a formal epsilon-delta proof (which I have not provided, but which would use the ideas below).
Key idea: the fact that $−3≤f(x)≤5$ (always) means that the function is bounded.  In particular, when $x$ is near $2$, you can always count on $f(x)$ to be some reliable number that neither gets very small nor very big. (Note that "small" and "big" here are relative terms; an upper bound of $1000000$ would be just as good as the upper bound of $5$ that the problem gives you, and a lower bound of $-1000000$ would be just as good as $-3$). 
Consider the product $(x-2)^2 f(x)$ as $x$ approaches $2$.  As $x$ approaches $2$, $(x-2)^2$ approaches $0$. As discussed earlier, $f(x)$ will stay nicely between $-3$ and $5$.  Therefore we are multiplying some arbitrarily small number  by some nicely predictable number that does not tend towards positive or negative infinity. So the product can be made arbitrarily small; i.e., it approaches $0$.
Using the fact that limits are additive, we see easily that $(x-2)^2 f(x)+1$ approaches $1$ as $x$ approaches $0$.
A: $$-3\le f(x)\le 5$$
$$-3(x-2)^2 \le (x-2)^2f(x)\le 5(x-2)^2$$
$$1-3(x-2)^2 \le (x-2)^2f(x)+1\le 1+5(x-2)^2$$
When $x\to2$, both $1-3(x-2)^2$ and $1+5(x-2)^2$ go to $1$.
Therefore, the required limit is $1$.
