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

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.

## closed as off-topic by Namaste, Xam, A. Goodier, Saad, ncmathsadistMar 24 '18 at 1:14

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• I doubt that the limit given in the title is of any use. – Yves Daoust Mar 23 '18 at 20:13
• I guess you use L'Hospital? – John Glenn Mar 23 '18 at 20:13

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?

• thank you! seems so simple now. but the way you factorized the numerator is to me not obvious at all. do you have any tips on how to do that? – cabralpinto Mar 23 '18 at 20:32
• The numerator is a quadratic polynomial with respect to $\sin x$ – Robert Z Mar 23 '18 at 20:41
• I feel dumb now :) thank you very much – cabralpinto Mar 23 '18 at 20:51

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 $2\cos(3x/2-\pi/4)$, $$\lim_{x\to-\pi/6}\cot\Bigl(\frac{x}{2}+\frac{\pi}{4}\Bigr)=\cot\frac{\pi}{6} =\sqrt{3}$$

• +1 - I like this approach a lot more personally, because it relies only on familiar identities and the common factor is immediate, rather than the little bit of magic that it feels like the double-angle identities require. – Steven Stadnicki Mar 23 '18 at 22:58

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