Limit-taking: Is this valid Is the following limit taking right? I am always confused as to when we are allowed to take the term-by-term limits then combine them as the correct full limit, sometimes term-by-term limit-taking doesn't give the right "full" limit... 
$$\lim\limits_{\epsilon\to0} {cf(x)f(x+\epsilon)\over c+\epsilon}= f^2(x)$$
perhaps I need to say that $f$ is continuous?
Thanks.
 A: Certainly you need that $f$ is continuous.  If not, there could be values of $f(x+\epsilon)$ very different from $f(x)$ even for very small $\epsilon$.  If $f$ is continuous this is correct.  Do you have the theorem that the limit of a product is the product of the limits?  You have $\lim\limits_{\epsilon\to0} {cf(x)f(x+\epsilon)\over c+\epsilon}=f(x)\lim\limits_{\epsilon\to0} (\frac{c}{c+\epsilon})f(x+\epsilon)$
A: If $c=0$ or $f(x)=0$, then the limit exists and is equal to $0$ regardless of $f(x)$.
If $c\neq 0$ and $f(x)\neq 0$, then the limit exists if and only if the limit
$$\lim_{\epsilon\to 0}f(x+\epsilon)$$
exists.
If $\lim\limits_{\epsilon\to 0}f(x+\epsilon)=L$, then
$$\lim_{\epsilon\to 0}\frac{cf(x)f(x+\epsilon)}{c+\epsilon} = cf(x)\left(\lim_{\epsilon\to 0}\frac{1}{c+\epsilon}\right)\left(\lim_{\epsilon\to 0}f(x+\epsilon)\right) = \frac{cf(x)}{c}L = f(x)L.$$
Conversely, if the limit you want exists, then so does the limit of
$$\left(\frac{c+\epsilon}{cf(x)}\right)\left(\frac{cf(x)f(x+\epsilon)}{c+\epsilon}\right) = f(x+\epsilon)$$
(a product of two functions that have a limit has itself a limit, equal to the product of the limits), and so the limit will necessarily equal $f(x)L$ as above.
Now, by definition of continuity, $L=\lim\limits_{\epsilon\to 0}f(x+\epsilon)=f(x)$ if and only if $f$ is continuous at $x$. 
So the limit equals $(f(x))^2$ if and only if either $f$ is $0$ at $x$; or if $c\neq 0$ and $f$ is continuous at $x$.
If you want the limit to equal $(f(x))^2$ for all $x$, then this holds if and only if $c\neq 0$ and $f(x)$ is continuous everywhere; or if $c=0$ and $f(x)=0$ for all $x$. 
