Use Continuity to evaluate the limit

[zonova]

Hello everyone, here's my problem.

Use continuity to evaluate the limit:
lim sin(x + sin(x)) , as x approaches pi

Now, i know that lim sin(x + sin(x)) = 0. However, i don't understand what it means to use Continuity to evaluate this limit. What does it want me to say? Should i go through the "3 step" for continuity? If i do the 3-step, is it assuming that i'm looking to see if the function is continuous at pi? I am simply confused as to what it is asking me to do

Also, i know that this question was answered before, a few years ago. It still leaves me confused on what to do, so i would appreciate any help

 

[HallsofIvy]

If the problem says use continuity, then, no, you are being asked to prove that sin(x) is continuous. It is saying that you may assume that you have already proved that sin(x) is continuous for all x.

Now, do you know what "continuous" means?? The simplest way or putting it, without a lot of technical points, is that a function, f(x), is continuous at x= a if and only if \(\displaystyle \lim_{x\to a} f(x)= f(a)\). Because, of course, \(\displaystyle a= \lim_{x\to a} x\), that can also be written in the very useful form \(\displaystyle \lim_{x\to a}f(x)= f(\lim_{x\to a} x)\).

Here, that means that \(\displaystyle \lim_{x\to \pi} sin(x+ sin(x))= sin(\lim_{x\to \pi}(x+ sin(x))\). Can you complete that?

 

[zonova]

Please correct me if i'm wrong in any way, as i'm still very new to Calculus.

So, if i have sin(lim_x->pi(x + sin(x)), i think the next step would be to split that up into:

sin(lim_x->pi(x) + lim_x->pi(sin(x)))

Which is sin(pi + sin(pi)) = sin(pi + 0) = sin(pi) = 0


Sorry, i haven't figured out how to use the symbols on this site properly.

 

[HallsofIvy]

Yes, that's right- saying that a function, f, is "continuous" at x= a means simply that the limit "as x goes to a" is exactly the same as f(a). So the limit as x goes to \(\displaystyle \pi\) of sin(x+ sin(x)) is just \(\displaystyle sin(\pi+ sin(\pi))= sin(\pi)= 0\).

 

[zonova]

Sorry, i think it reposted what you had posted earlier.

Edit: Oh, you changed it right as i posted that, my mistake Thanks for all of the help. I just have another question, in a bit of the same light. It's still dealing with using continuity. The problem is:

lim_x->2(x³-3x+1)^-3

Now, i know that since the function is rational, it is continuous on it's domain. However, i have no idea how to calculate its domain. Is it necessary for me to specify the domain? Thanks again for your help

 

[pka]

Use continuity to evaluate the limit:
lim sin(x + sin(x)) , as x approaches pi

I think that you are being asked to use properties of continuous functions.

If each of \(\displaystyle f~\&~g\) is a continuous function then each of \(\displaystyle f+g~\&~f\circ g \) is continuous.

We know that each of \(\displaystyle x~\&~\sin(x) \) is a continuous function so \(\displaystyle \sin(x+\sin(x) \) is continuous. WHY?

How do we use that to find \(\displaystyle \displaystyle{\lim _{x \to \pi }}\sin \left( {x + \sin (x)} \right)~?\)

 

[zonova]

Well, through what HallsofIvory explained, it basically ended up being that i needed to plug in pi for each x. That brought me to it being 0. Thanks for the help with that problem Would you be able to assist me with the other one as well? I described it in the post right above yours

 

[pka]

I just have another question, in a bit of the same light. It's still dealing with using continuity. The problem is:
lim_x->2(x³-3x+1)^-3

Now, i know that since the function is rational, it is continuous on it's domain. However, i have no idea how to calculate its domain.

There is no need to find the domain. In fact is difficult to do.
BUT you do want to know if \(\displaystyle 2\) is in that domain? Yes it is.
So \(\displaystyle \displaystyle{\lim _{x \to 2}}\frac{1}{{{{\left( {{x^3} - 3x + 1} \right)}^3}}} =\frac{1}{{{{\left( {{2^3} - 3 \cdot 2 + 1} \right)}^3}}}\)