Question

Question #99ee1

Answer 1

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

`lim_{x\to x_0} f(x)=f(x_0)`

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\to x_0} f(x)=f(lim_{x\to x_0}x)`

This given, your solution is thus

`\lim_{x\to\pi} \sin(x+\sin(x))=`
`\sin(\lim_{x\to\pi} (x+\sin(x)))`

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

`\sin(\lim_{x\to\pi} (x+\sin(x)))=`
`\sin(\pi+\sin(\lim_{x\to\pi} x)=`
`\sin(\pi+\sin(\pi))`

And since `\sin(\pi)=0`, we have

`\sin(\pi+\sin(\pi))=`
`\sin(\pi+0)=\sin(\pi)=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(x)=0`
graph{x+sin(x) [-10, 10, -5, 5]}