Question #99ee1

↳Redirected from "Question #99ee1"
1 Answer
KillerBunny
Mar 24, 2015

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]}

Related questions
See all questions in Limits Involving Trigonometric Functions
Impact of this question
12709 views around the world
You can reuse this answer
Creative Commons License