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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!

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Let $f(x):(a,b)\to R$ and $g(x):(a,b)\to R$ be continuous at the point $x_o$ $ϵ$ $(a,b)$. Then $f(x)/g(x)$ is continuous at the point $x_o$ $ϵ$ $(a,b)$ , for $g(x_0)$ different from zero.

Proof:

Knowing that $f$ and $g$ are continuous at the point $x_o$ $ϵ$ $(a,b)$ 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:

$|f(x)/g(x)-f(x_0)/g(x_0)|=|f(x)g(x_0)-g(x)f(x_0)|/|g(x)g(x_0)|=|f(x)g(x_0)-f(x_0)g(x_0)+f(x_0)g(x_0)-g(x)f(x_0)|/|g(x)g(x_0)|\leq(|g(x_0)||f(x)-f(x_0)|+|f(x_0)||g(x)f(x_0)|)|g(x)g(x_0)|$

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

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

so we get:

$|f(x)/g(x)-f(x_0)/g(x_0)|\leq(|g(x_0)||f(x)-f(x_0)|+|f(x_0)||g(x)f(x_0)|)|g(x)g(x_0)|<2(|g(x_0)||f(x)-f(x_0)|+|f(x_0)||g(x)-g(x_0)|)/|g(x_0)|^2=2|f(x)-f(x_0)|/|g(x_0)|+2|f(x_0)||g(x)-g(x_0)|/|g(x_0)|^2$

Now , let $M=\max\{2/|g(x_0)|,2|f(x_0)|/|g(x_0)|^2\}$

and let $\epsilon_1=\epsilon_2=\epsilon/2|M|$ and let $\delta=min\{\delta_1,\delta_2\}$

We get:

$|f(x)/g(x)-f(x_0)/g(x_0)|\leq 2|f(x)-f(x_0)|/|g(x_0)|+2|f(x_0)||g(x)-g(x_0)|/|g(x_0)|^2<|M|(|f(x)-f(x_0)|+|g(x)-g(x_0)|)<|M|\epsilon/2|M|+|M|\epsilon/2|M|<\epsilon$

Proved.

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  • $\begingroup$ You typed this all in 1 minute? Wow...? $\endgroup$ Mar 6 '18 at 17:44
  • $\begingroup$ @ThePhenotype I found it . $\endgroup$ Mar 6 '18 at 17:48
  • $\begingroup$ @MathsSurvivor Thank you ! $\endgroup$
    – user414790
    Mar 6 '18 at 17:52
  • $\begingroup$ Ah okay, could you give me the source? $\endgroup$ Mar 6 '18 at 18:02
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    $\begingroup$ @ThePhenotype sure jpetrie.net/2015/05/07/… I just made a little change at the end. $\endgroup$ Mar 6 '18 at 18:17
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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 $\delta$)

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