Evaluating $ \lim _{x\to \:-\pi /6}\frac{\cos 2x+\sin x}{\sin 2x+\cos x}$

Question

Here's the problem:

$$ \lim _{x\to \:-\frac{\pi }{6}}\frac{\cos \left(2x\right)+\sin \left(x\right)}{\sin \left(2x\right)+\cos \left(x\right)}$$

I'm pretty sure I am supposed to use the notable limit

$$\lim_{x\to 0} \frac{\sin x}{x} = 1$$

given the context of what I am studying. I've tried multiple ways and just kept getting stuck in indeterminations. Please help.

Answer 1

You don't need any special limit here. Just note that $$\frac{\cos \left(2x\right)+\sin \left(x\right)}{\sin \left(2x\right)+\cos \left(x\right)}=\frac{1-2\sin^2(x)+\sin \left(x\right)}{2\sin(x)\cos(x)+\cos(x)}=\frac{(2\sin(x)+1)(1-\sin(x))}{(2\sin(x)+1)\cos(x)}.$$ Can you take it from here?

Answer 2

Take the derivative of the numerator and denominator so you get: $$\lim _{x\to \:-\frac{\pi }{6}}\left(\frac{\cos \left(2x\right)+\sin \left(x\right)}{\sin \left(2x\right)+\cos \left(x\right)}\right)=\frac{\cos (-\frac{\pi }{6})-2 \sin (2 (-\frac{\pi }{6}))}{2 \cos (2 (-\frac{\pi }{6}))-\sin (-\frac{\pi }{6})}=\sqrt3$$

Answer 3

Not that this method, in this case, is more efficient than factorization, but it can come handy in other situations.

Note that $\sin x=\cos(\frac{\pi}{2}-x)$ and, with $$ \cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} $$ the numerator becomes $$ 2\cos\Bigl(\frac{x}{2}+\frac{\pi}{4}\Bigr)\cos\Bigl(\frac{3x}{2}-\frac{\pi}{4}\Bigr) $$ Similarly, $\cos x=\sin(\frac{\pi}{2}-x)$ and, with $$ \sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} $$ the denominator becomes $$ 2\sin\Bigl(\frac{x}{2}+\frac{\pi}{4}\Bigr)\cos\Bigl(\frac{3x}{2}-\frac{\pi}{4}\Bigr) $$ Thus you get, after factoring out $2\cos(3x/2-\pi/4)$, $$ \lim_{x\to-\pi/6}\cot\Bigl(\frac{x}{2}+\frac{\pi}{4}\Bigr)=\cot\frac{\pi}{6} =\sqrt{3} $$