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.
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$\begingroup$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 $\;g(x)=1\;$ ...
$\endgroup$ 2 $\begingroup$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 $.
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