# 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.

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Is $c$ nonzero? – J. M. Nov 4 '11 at 14:10
@J.M.: Yes, $c\neq 0$. – kid Nov 4 '11 at 14:12
In most natural situations, term by term limit taking, when it gives an answer, gives the right answer. Note that important cases that you have been introduced to, such as $\displaystyle\lim\frac{\sin x}{x}$, are consistent with the above claim. If we take the limit of the top, the limit of the bottom, we do not get a wrong answer. We just get the expression $\frac{0}{0}$, which is utterly unhelpful. – André Nicolas Nov 4 '11 at 16:29

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 $\lim\limits_{\epsilon\to0} {cf(x)f(x+\epsilon)\over c+\epsilon}=f(x)\lim\limits_{\epsilon\to0} (\frac{c}{c+\epsilon})f(x+\epsilon)$

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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 $x$.

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