Quotient rule for limits of functions

Question

Suppose that, $\lim\limits_{x\to x_0}f(x)$ and $\lim\limits_{x\to x_0}g(x)$ exist and $\lim\limits_{x\to x_0}g(x)\neq0$. Then, prove that $$\lim\limits_{x\to x_0}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\to x_0}f(x)}{\lim\limits_{x\to x_0}g(x)}$$

Answer 1

Let $K$ and $L$ be the limits of $f$ and $g$ respectively. Let $\epsilon > 0 $ and choose $\delta > 0 $ such that for $|x-x_0| < \delta$ we have $|f(x) - K| < \epsilon$ and $|g(x) - L| < \epsilon.$ We can also conveniently suppose that in this range $|g(x)| \geq |L|/2 .$ Then

\begin{align} |f(x)/g(x) - K/L| &= \bigg|\frac{f(x)L - Kg(x)}{Lg(x)}\bigg| \\ &= \bigg|\frac{f(x)L - KL + KL - Kg(x)}{Lg(x)}\bigg| \\ &= \bigg|\frac{L(f(x) - K) + K(L - g(x))}{Lg(x)}\bigg| \\ &< \epsilon(L+K)/L|g(x)|\\ &\leq \epsilon2(L+K)/L^2. \end{align} Of course if you want all of this to be neatly less than $\epsilon$ then you could change your values of $\delta$ accordingly.