[tex]\displaystyle\lim_{x\to\pi/5}\sin(x+\sin x)=\sin\left(\lim_{x\to\pi/5}(x+\sin x)\right)[/tex]
[tex]=\displaystyle\sin\left(\lim_{x\to\pi/5}x+\lim_{x\to\pi/5}\sin x\right)[/tex]
[tex]=\displaystyle\sin\left(\dfrac\pi5+\sin\left(\lim_{x\to\pi/5}x\right)\right)[/tex]
[tex]=\displaystyle\sin\left(\dfrac\pi5+\sin\dfrac\pi5\right)[/tex]
You can stop right there, or you can try finding the exact value of [tex]\sin\dfrac\pi5[/tex].
Recall DeMoivre's theorem:
[tex](\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta[/tex]
This means when [tex]n=5[/tex], the imaginary part of the expansion of the left side will give you an expanded form of [tex]\sin5\theta[/tex] in terms of powers of [tex]\sin\theta[/tex]. You have
[tex]\mathrm{Im}\left[(\cos\theta+i\sin\theta)^5\right]=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta[/tex]
[tex]=5\sin\theta-20\sin^3\theta+16\sin^5\theta[/tex]
where the last equality comes from the fact that [tex]\cos^2\theta+\sin^2\theta=1[/tex]. So
[tex]\sin5\theta=5\sin\theta-20\sin^3\theta+16\sin^5\theta[/tex]
Now, setting [tex]\theta=\dfrac\pi5[/tex], you get
[tex]\sin\pi=5\sin\dfrac\pi5-20\sin^3\dfrac\pi5+16\sin^5\dfrac\pi5[/tex]
[tex]0=\sin\dfrac\pi5\left(5-20\sin^2\dfrac\pi5+16\sin^4\dfrac\pi5\right)[/tex]
Clearly, [tex]\sin\dfrac\pi5\neq0[/tex], so you're left with the quartic equation
[tex]0=5-20\sin^2\dfrac\pi5+16\sin^4\dfrac\pi5[/tex]
Applying the quadratic formula gives a solution of
[tex]\sin^2\dfrac\pi5=\dfrac{20\pm\sqrt{80}}{32}=\dfrac{20\pm4\sqrt5}{32}=\dfrac{5\pm\sqrt5}8[/tex]
Since [tex]\sin^2\dfrac\pi4=\dfrac12[/tex], we should expect [tex]\sin^2\dfrac\pi5[/tex] to be smaller, which means we take the positive root because [tex]\dfrac58>\dfrac12[/tex], and adding a positive number would make this larger. So,
[tex]\sin^2\dfrac\pi5=\dfrac{5-\sqrt5}8[/tex]
which means
[tex]\sin\dfrac\pi5=\pm\sqrt{\dfrac{5-\sqrt5}8}[/tex]
but we also expect this number to be positive, so we ignore the negative root and end up with
[tex]\sin\dfrac\pi5=\sqrt{\dfrac{5-\sqrt5}8}[/tex]
So the limit is
[tex]\sin\left(\dfrac\pi5+\sqrt{\dfrac{5-\sqrt5}8}\right)[/tex]
Now, there's no reason to expect this to have a simpler form, so we can stop here. (Perhaps this answer is overkill, but if you didn't know this stuff, it doesn't hurt to learn it.)
