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Consider the following limit:

$$L=\lim_{x\to a}\left(\frac{f(x)g(x)}{p(x)q(x)}\right)~,~a\in\Bbb{R}$$

I know that $\displaystyle\lim_{x\to a}\dfrac{f(x)}{p(x)}$ and the limit $L$ both exist finitely. Now, I was wondering whether we could separate the limits in $L$ like this using the info we have till now?

$$L=\left(\lim_{x\to a}\frac{f(x)}{p(x)}\right)\left(\lim_{x\to a}\frac{g(x)}{q(x)}\right)$$

Note: At $x=a$, the body of the limit $L$ is undefined (not in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$).

Actually, I wanted to know this to prove another result that uses the following idea. I was trying at that other problem using proof by contradiction and I need to know whether this method of separation works to conclude my proof of that problem. I don't want to post that other problem, so I'm posting this general question.

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  • $\begingroup$ You can separate limits like that provided that you know that all the limits in the separated version exist and are finite. $\endgroup$ – TravisJ Apr 2 '15 at 19:41
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    $\begingroup$ @TravisJ, the thing is, I wanted to employ this in a proof by contradiction where I'll later prove that $\displaystyle\lim_{x\to a}\frac{g(x)}{q(x)}$ doesn't exist finitely and hence conclude my proof. So, you see, we can't use that. $\endgroup$ – learner Apr 2 '15 at 19:43
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    $\begingroup$ If the limit of $f(x)/p(x)$ is $0$, strange things can happen. $\endgroup$ – André Nicolas Apr 2 '15 at 19:44
  • $\begingroup$ If the body of the limit is undefined in a diverging manner (Such as $\frac\infty{c}$ or $\frac c0$, the equality $L=\ldots$ must fail because you just said that the RHS is undefined. So what is it exactly that you know of $\frac{f(x)g(x)}{p(x)q(x)}$? $\endgroup$ – AlexR Apr 2 '15 at 19:46
  • $\begingroup$ @AndréNicolas, Assume that the limit of $f(x)/p(x)$ is $1$ for simplicity's sake. That assumption would work for me since the problem I'm dealing with has that. $\endgroup$ – learner Apr 2 '15 at 19:47
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Note that the limits $$L_a = \lim_{x\to a} \frac{f(x)}{p(x)}$$ and $$L = \lim_{x\to a} \frac{f(x)g(x)}{p(x)q(x)}$$ exist and are finite. Additionally $L_a = 1 \ne 0$ so we know that $$L_b = \frac L{L_a} = \frac{\lim\limits_{x\to a}\frac{f(x)g(x)}{p(x)q(x)}}{\lim\limits_{x\to a}\frac{f(x)}{p(x)}} = \lim_{x\to a}\frac{\frac{f(x)g(x)}{p(x)q(x)}}{\frac{f(x)}{p(x)}} = \lim_{x\to a} \frac{g(x)}{q(x)}$$ exists and is finite. This allows us to separate the limits as you want to.

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