Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

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

share|improve this answer

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 $x$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.