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In real analysis, we have been asked to finish a proof of the quotient rule for limits (Given that $f(x)$ approaches $L$ and $g(x)$ approaches M as $x$ approaches a, prove that $\frac{f(x)}{g(x)}$ approaches $\frac{L}{M}$. I know that I could rewrite the quotient as multiplication and prove it that way but that is not the way the proof we are completing starts off. Here is how the proof in the book starts:

We showed in a previous exercise that $|g(x)| > \frac{|M|}{2}$ for all $x$ belonging to $D$ near $a$. In particular, $g(x)$ does not equal $0$ for all $x$ belonging to $D$ near $a$. Consequently, $\frac{f(x)}{g(x)}$ makes sense near $a$. Let $\epsilon > 0$ be given. The proof is based on the triangle inequality: \begin{align*} \left|\frac{f(x)}{g(x)} - \frac{L}{M}\right| &= \left|\frac{f(x)}{g(x)} - \frac{L}{g(x)} + \frac{L}{g(x)} - \frac{L}{M}\right|\\ & = \left|\frac{f(x)}{g(x)} - \frac{L}{g(x)} + L \frac{M - g(x)}{Mg(x)}\right| \\ &\leq \left|\frac{1}{g(x)}\right| |f(x) - L| + \left|\frac{L}{M}\right| \left|\frac{1}{g(x)}\right| |M - g(x)|. \end{align*} We are expected to pick up the proof at this point.

I was thinking that since I want everything above $< \frac{\epsilon}{2} + \frac{\epsilon}{2}=\epsilon$, that I would construct $\epsilon_f$ and $\epsilon_g$ based on $\epsilon$. So, I said we want $|\frac{1}{g(x)}| |f(x) - L| < \epsilon/2$ and because the limit approaches $L$ end up with $\epsilon_f$ equals $\frac{\epsilon|M|}{4}$. I intended to do the same for the g part; however, when I go to define $\epsilon_g$ it involves $M$. Am I allowed to use $M$ in that definition for $\epsilon_g$? And if not, any ideas on where to go from here?

Thank you!!

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  • $\begingroup$ For MathJax: meta.math.stackexchange.com/q/5020 $\endgroup$ – user142971 Sep 16 '16 at 19:01
  • $\begingroup$ Yes, you want to use a similar idea to find $\epsilon_{g}$ -- you want to have $\frac{|L|}{|M|}\frac{2}{|M|}|g(x)-M|<\frac{\epsilon}{2}$. (Notice that this is automatically true if $L=0$) $\endgroup$ – user84413 Sep 16 '16 at 19:26
  • $\begingroup$ When doing that though, can my $\epsilon_g$ involve M or not? I was thinking it could not because M is involved in the limit with g(x). If it can, I know how to do it. I had done it so that $\epsilon_g$ = $\frac{\epsilon M^2}{4|L|}$. $\endgroup$ – user369580 Sep 16 '16 at 19:30
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Since $|g(x)|>|M|/2$, you know that $$\left|\dfrac{1}{g(x)}\right|<\dfrac{2}{|M|}$$ if $x$ is near $a$. How small should $|f(x)-L|$ be so that $$\left|\dfrac{1}{g(x)}\right|\,|f(x)-L|<\dfrac{\varepsilon}{2}$$ be satisfied? Remember that you can use the definition of the limit with any number instead of $\varepsilon$.

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  • $\begingroup$ I am able to do that part, the problem is when working with the other half of the expression. I satisfied the inequality you discussed in that last paragraph. The part involving g(x) is the problem because I cannot use M when talking about my epsilon for that part right because M is what is approaching? $\endgroup$ – user369580 Sep 16 '16 at 19:28
  • $\begingroup$ The second part is not any different. There's just an extra factor $|L|/|M|$ which is a constant (possibly 0). You have to make $|g(x)-M|$ small enough so that the product of these three things is smaller than $\varepsilon/2$. You're absolutely allowed to use $M$ in that definition, it is a given constant in this context. $\endgroup$ – Olivier Moschetta Sep 16 '16 at 19:30
  • $\begingroup$ Okay, perfect - thank you so much!! $\endgroup$ – user369580 Sep 16 '16 at 19:31

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