# How to prove that f is continuous?

## $f , g : R \to R$ with $g \left(x\right) = f \left(f \left(x\right)\right) + {e}^{x}$ ,xεR supposed $f$, $g$ are strictly increasing and for x_0εR, ${\lim}_{x \rightarrow {x}_{0}} f \left(x\right) = l$

Dec 5, 2017

${\lim}_{x \rightarrow {x}_{0}} f \left(x\right) = f \left({x}_{0}\right) = L$

#### Explanation:

${\lim}_{x \rightarrow {x}_{0}} f \left(x\right) = L$ $\iff$${\lim}_{x \rightarrow {x}_{0}^{+}} f \left(x\right)$=${\lim}_{x \rightarrow {x}_{0}^{-}} f \left(x\right) = L$

• If $x <$${x}_{0}$ $\implies$ $x \to {x}_{0}^{-}$ $\implies$$f \left(x\right)$$<$$f \left({x}_{0}\right)$ , because $f$ strictly increasing

so $\implies$ ${\lim}_{x \rightarrow {x}_{0}^{-}} f \left(x\right)$ $\le$$f \left({x}_{0}\right)$ $\left(1\right)$

• If $x >$${x}_{0}$ $\implies$ $\implies$ $x \to {x}_{0}^{+}$ $\implies$$f \left(x\right)$$>$$f \left({x}_{0}\right)$ , because $f$ strictly increasing

so $\implies$ ${\lim}_{x \rightarrow {x}_{0}^{+}} f \left(x\right)$ $\ge$$f \left({x}_{0}\right)$ $\left(2\right)$

$L \le f \left({x}_{0}\right)$
&
$L \ge f \left({x}_{0}\right)$

$\implies$ $f \left({x}_{0}\right) = L$