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I hope you guys are enjoying your weekend. I have a question about limits.

This homework problem asks me to use continuity to evaluate this limit, I would like to double-check that I have following the right procedure.

The problem is as follows:

$$\lim_{x\to \pi}\sin(x + \sin x)$$

I break the problem up into two seperate limits:

$$\lim_{x\to \pi}\sin x + \lim_{x\to \pi}\sin(\sin x)$$

Because $\sin x$ is continuous in its domain and its domain includes all real numbers, both limits are continuous and have a domain that includes all real numbers.

I can therefore plug in $\pi$ and conclude that the limit is $0$.

Is my methodology correct or am I making a mistake?


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it's not true that $\sin(x+y) = \sin(x) + \sin(y)$ – mm-aops Jun 23 '14 at 1:04
To follow up with @mm-aops' comment: what is true is that $\sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(y)$. – Cameron Williams Jun 23 '14 at 1:10
up vote 5 down vote accepted

Using continuity to evaluate the problem means that you can use the following fact (assuming you proved it in class, not sure what else your teacher might have been asking) for a continuous function $f:\mathbb{R} \to \mathbb{R}$: $$\lim_{x\to a} f(x)=f(\lim_{x\to 0} x).$$

So $\lim_{x\to \pi}\sin(x+\sin x)=\sin(\lim_{x\to \pi}x+\sin x)=\sin(\pi+0)=0$

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You can make sine, cosine, etc align upright by typing \sin, \cos, etc. It makes your math look a little bit nicer. :) – Cameron Williams Jun 23 '14 at 1:12
You are right! Just fixed it – Sixter Jun 23 '14 at 1:14
This is great, thanks! – Irresponsible Newb Jun 23 '14 at 1:19

For a function $f:\mathbf{R}\to\mathbf{R}$ we have that $f$ is continuous at every point in $\mathbf{R}$ if $f$ is "differentiable", that is, able to be differentiated. In this case, if we take $f(x) = \sin(x + \sin x)$, then we have that

$$\frac{\mathrm{d}f}{\mathrm{d}x}=\cos(x+\sin x)(1+\cos x),$$

and so we may justly claim that $f$ is continuous over $\mathbf{R}$. Since

$$\lim_{x\to a}h(x)=h(a),$$

for some continuous function $h$, we then have

$$\lim_{x\to \pi}\sin(x+\sin x)=\sin(\pi+\sin\pi)=0.$$

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