# Show a rational function is continuous

This question was is from a real analysis test I wrote last week, and I don't have any idea how to solve it. It's nowhere in my notes as an example either.

Let $f(x)=\frac 1 {1+x}, x \in [-\frac 12;\frac 12]$

$c$ is any real constant and it is given that $g(x)=c$, and $h(x)=x$ are continuous on $\Bbb R$

Show that $f$ is continuous on $[-\frac 34;-\frac 12]$

I don't really have any ideas using what is given. It feels like I am missing a tool here.

Does anyone have a solution? The context of this test was sequences and series of functions.

• If your givens ar so basic, you should also include that the reciprocal is continuous in its domain of definition. - On the other hand, $f$ cannot be continuous on $[-\frac34,-\frac12]$ because it isn't even defined there (exept at $-\frac12$) Jun 1 '14 at 12:56
• How come $\;f(x):=\frac1{1+x}\;$ isn't defined, and thus not continuous, at that interval, @HagenvonEitzen ? Oh, you mean because of the very first given expression? Jun 1 '14 at 12:57
• @HagenvonEitzen it always irks me about tests I have received that things like this aren't properly defined, so I am forced to assume it to get anywhere... but then assume one thing the lecturer was not expecting and it is marked incorrect. Jun 1 '14 at 13:00

I think they mean:

$$g,h\;\;\text{continuous}\implies g+h \;\;\text{continuous}\;\implies\frac1{g+h}\;\text{continuous whenever}\;g+h\neq 0$$

and now just choose $\;g(x)=1\;$ ...

• That makes a lot of sense, but it leaves me wondering why we were given the relatively obvious fact that $c$ is continuous but not that $(g+h) continuous \implies \frac 1 {g+h} continuous$ Jun 1 '14 at 13:04
• Well, it is a theorem @Matt : prove it, or you can check it in any decent calculus book. Jun 1 '14 at 13:06

This is just a composition of continuous functions.

If you want to do a complete proof, go back to the definition.

Let $x_0 \in [-1/2,1/2]$ and $\epsilon > 0$, then for any $x\in [-1/2,1/2]$,

$|f(x)-f(x_0)| = |x-x_0|\dfrac{1}{|1+x||1+x_0|}$

In particular $x,x_0 \in [-1/2,1/2]$ implies $|1+x||1+x_0| \geq \dfrac{1}{4}$. So for any $x\in [-1/2,1/2]$ satisfying $|x-x_0| \leq \dfrac{\epsilon}{4}$ we get

$|f(x)-f(x_0)| = \dfrac{|x-x_0|}{|1+x||1+x_0|} \leq 4|x-x_0| \leq \epsilon$.