# Question 99ee1

Mar 24, 2015

A function $f$ is continuous if the following holds:

${\lim}_{x \setminus \to {x}_{0}} f \left(x\right) = f \left({x}_{0}\right)$

Which means that, if you consider the limit of $x$ approaching some value, the limit will be exactly the function evaluated in that value.

Visually, this means that you can bring the limit "inside" the function, in this sense:

${\lim}_{x \setminus \to {x}_{0}} f \left(x\right) = f \left({\lim}_{x \setminus \to {x}_{0}} x\right)$

This given, your solution is thus

$\setminus {\lim}_{x \setminus \to \setminus \pi} \setminus \sin \left(x + \setminus \sin \left(x\right)\right) =$
$\setminus \sin \left(\setminus {\lim}_{x \setminus \to \setminus \pi} \left(x + \setminus \sin \left(x\right)\right)\right)$

Now, of course $\setminus {\lim}_{x \setminus \to \setminus \pi} x = \setminus \pi$, and applying one more time this continuity rule to the inner sine function:

$\setminus \sin \left(\setminus {\lim}_{x \setminus \to \setminus \pi} \left(x + \setminus \sin \left(x\right)\right)\right) =$
\sin(\pi+\sin(\lim_{x\to\pi} x)=#
$\setminus \sin \left(\setminus \pi + \setminus \sin \left(\setminus \pi\right)\right)$

And since $\setminus \sin \left(\setminus \pi\right) = 0$, we have

$\setminus \sin \left(\setminus \pi + \setminus \sin \left(\setminus \pi\right)\right) =$
$\setminus \sin \left(\setminus \pi + 0\right) = \setminus \sin \left(\setminus \pi\right) = 0$

Here're the graph of the function, showing both that the function is continuous (even if it's not a proof of course), and that $f \left(x\right) = 0$
graph{x+sin(x) [-10, 10, -5, 5]}