\(\displaystyle\lim_{x\to\pi/5}\sin(x+\sin x)=\sin\left(\lim_{x\to\pi/5}(x+\sin x)\right)\)
\(=\displaystyle\sin\left(\lim_{x\to\pi/5}x+\lim_{x\to\pi/5}\sin x\right)\)
\(=\displaystyle\sin\left(\dfrac\pi5+\sin\left(\lim_{x\to\pi/5}x\right)\right)\)
\(=\displaystyle\sin\left(\dfrac\pi5+\sin\dfrac\pi5\right)\)
You can stop right there, or you can try finding the exact value of \(\sin\dfrac\pi5\).
Recall DeMoivre's theorem:
\((\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta\)
This means when \(n=5\), the imaginary part of the expansion of the left side will give you an expanded form of \(\sin5\theta\) in terms of powers of \(\sin\theta\). You have
\(\mathrm{Im}\left[(\cos\theta+i\sin\theta)^5\right]=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta\)
\(=5\sin\theta-20\sin^3\theta+16\sin^5\theta\)
where the last equality comes from the fact that \(\cos^2\theta+\sin^2\theta=1\). So
\(\sin5\theta=5\sin\theta-20\sin^3\theta+16\sin^5\theta\)
Now, setting \(\theta=\dfrac\pi5\), you get
\(\sin\pi=5\sin\dfrac\pi5-20\sin^3\dfrac\pi5+16\sin^5\dfrac\pi5\)
\(0=\sin\dfrac\pi5\left(5-20\sin^2\dfrac\pi5+16\sin^4\dfrac\pi5\right)\)
Clearly, \(\sin\dfrac\pi5
eq0\), so you're left with the quartic equation
\(0=5-20\sin^2\dfrac\pi5+16\sin^4\dfrac\pi5\)
Applying the quadratic formula gives a solution of
\(\sin^2\dfrac\pi5=\dfrac{20\pm\sqrt{80}}{32}=\dfrac{20\pm4\sqrt5}{32}=\dfrac{5\pm\sqrt5}8\)
Since \(\sin^2\dfrac\pi4=\dfrac12\), we should expect \(\sin^2\dfrac\pi5\) to be smaller, which means we take the positive root because \(\dfrac58>\dfrac12\), and adding a positive number would make this larger. So,
\(\sin^2\dfrac\pi5=\dfrac{5-\sqrt5}8\)
which means
\(\sin\dfrac\pi5=\pm\sqrt{\dfrac{5-\sqrt5}8}\)
but we also expect this number to be positive, so we ignore the negative root and end up with
\(\sin\dfrac\pi5=\sqrt{\dfrac{5-\sqrt5}8}\)
So the limit is
\(\sin\left(\dfrac\pi5+\sqrt{\dfrac{5-\sqrt5}8}\right)\)
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.)
