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!!
Best Answer
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 $\varepsilon$.