Question

Use continuity to evaluate the limit. lim x→π 5 sin(x + sin x)

Answer 1

$\displaystyle\lim_{x\to\pi/5}\sin(x+\sin x)=\sin\left(\lim_{x\to\pi/5}(x+\sin x)\right)$
$=\displaystyle\sin\left(\lim_{x\to\pi/5}x+\lim_{x\to\pi/5}\sin x\right)$
$=\displaystyle\sin\left(\dfrac\pi5+\sin\left(\lim_{x\to\pi/5}x\right)\right)$
$=\displaystyle\sin\left(\dfrac\pi5+\sin\dfrac\pi5\right)$

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, $\sin\dfrac\pi5\neq0$, 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.)