Prove that if $f$ and $g$ are continuous functions then $f/g$ is also continuous

Question

I have to prove this, I know what does it mean for a function to be continuous using $\epsilon-\delta$ definition but yet I'm not being able to prove this one , I've searched on the internet but there's no proof for this one there are only proofs that sum or multiplication of continuous functions is continuous but there's no proof that dividing two continuous functions will give a continuous function. Any help will be appreciated.Thank you!

Answer 1

Let $f, g:E\to \mathbb{R}$ be continuous at the point $x_0 \in E$, suppose $g(x) \neq 0$ for all $x \in E$, then $\frac{f(x)}{g(x)}$ is continuous at the point $x_0 \in E$.

Proof:

Knowing that $f$ and $g$ are continuous at the point $x_0 \in E$ we have that:

1.For every $\epsilon_1>0$ there exists a $\delta_1>0$ such that for every $|x-x_0|<\delta_1$ $\implies$ $|g(x)-g(x_0)|<\epsilon_1$

2.For every $\epsilon_2>0$ there exists a $\delta_2>0$ such that for every $|x-x_0|<\delta_2$ $\implies$ $|f(x)|-f(x_0)|<\epsilon_2$

We need to show that $\lim_{x \to x_0} f(x)/g(x)=f(x_0)/g(x_0)$ we use the $\epsilon-\delta$ definition to prove it :

for every $\epsilon>0$ there exists a $\delta>0$ such that for every $x$ that satisfies the inequality $|x-x_0|<\delta$ implies that $|f(x)/g(x)-f(x_0)/g(x_0)|<\epsilon$

We have: \begin{align*} \left|\frac{f(x)}{g(x)}-\frac{f(x_0)}{g(x_0)}\right|&=\frac{|f(x)\cdot g(x_0)-g(x) \cdot f(x_0)|}{|g(x)\cdot g(x_0)|}\\ &=\frac{|f(x)\cdot g(x_0)-f(x_0) \cdot g(x_0)+f(x_0) \cdot g(x_0)-g(x) \cdot f(x_0)|}{|g(x) \cdot g(x_0)|} \\ &\leq \frac{|g(x_0)|\cdot |f(x)-f(x_0)|+|f(x_0)|\cdot |g(x) - g(x_0)|}{|g(x)\cdot g(x_0)|} \end{align*}

Let $\epsilon_1=\frac{|g(x_0)|}{2}>0$ then there exists a $\delta_1>0$ such that $|x-x_0|<\delta_1$ $\implies$ $|g(x_0)|-|g(x)|\leq|g(x)-g(x_0)|< \frac{|g(x_0)|}{2} = \epsilon_1$

$\implies$ $1/|g(x)|<2/|g(x_0)|$

so we get: \begin{align*} |\frac{f(x)}{g(x)}-\frac{f(x_0)}{g(x_0)}| &\leq \frac{|g(x_0)|\cdot |f(x)-f(x_0)|+|f(x_0)|\cdot |g(x) - g(x_0)|}{|g(x)\cdot g(x_0)|}\\ & < \frac{2\cdot \left(|g(x_0)|\cdot |f(x)-f(x_0)|+|f(x_0)|\cdot |g(x)-g(x_0)| \right)}{|g(x_0)|^2}\\ &=\frac{2\cdot |f(x)-f(x_0)|}{|g(x_0)|}+\frac{2\cdot |f(x_0)|\cdot |g(x)-g(x_0)|}{|g(x_0)|^2} \end{align*}

Now , let $M=\max\left\{\frac{2}{|g(x_0)|},\frac{2|f(x_0)|}{|g(x_0)|^2}\right\}>0$

and let $\epsilon_1=\epsilon_2=\frac{\epsilon}{2\cdot M}>0$ and let $\delta=min\{\delta_1,\delta_2\}>0$

We get: \begin{align*} \left|\frac{f(x)}{g(x)}-\frac{f(x_0)}{g(x_0)}\right| & < \frac{2\cdot |f(x)-f(x_0)|}{|g(x_0)|}+\frac{2\cdot |f(x_0)|\cdot |g(x)-g(x_0)|}{|g(x_0)|^2}\\ & <M \cdot \left(|f(x)-f(x_0)|+|g(x)-g(x_0)|\right)\\ & <M \cdot \frac{\epsilon}{2\cdot M}+M \cdot \frac{\epsilon}{2\cdot M}\\ &=\epsilon \end{align*}

Now, given $f, g: E \rightarrow \mathbb{R}$ are continuous at $x_0 \in E$ and $g(x) \neq 0$ for all $x \in E$, then $\frac{f}{g}: E \rightarrow \mathbb{R}$ is continuous at $x_0 \in E$.

Q.E.D.

Answer 2

Here's a quicker (and less rigorous) answer:

Because $g(x_0)$ is non-zero, set delta sufficiently small such that $|g(x)| > |g(x_0)|/2$ for all $|x_0 - \delta| < |x| < |x_0 + \delta|$. We can do this because $g$ is continuous and $g(x_0)$ is non-zero.

When you are given an $\epsilon > 0$, you can show that $$ |f/g (x) - f/g (x_0)| < \frac{|g(x_0)(f(x) - f(x_0))|}{|g(x)g(x_0)|} + \frac{|f(x_0)(g(x) - g(x_0))|}{|g(x)g(x_0)|} $$ and both the LH and RH terms can be made arbitrarily small by choosing a sufficiently small $\delta$, by A) Continuity of $f$ and $g$ at $x_0$, and B) The fact that the denominator is always greater than $|g(x_0)|^2 / 4$ (because of how we chose our $\delta$)