proving that $\lim_{x \to \pi} \frac{1+\cos x }{1+\cos 3x}=\frac{1}{9}$ without L'Hôpital.

Question

So basically I'm trying to show $$\lim_{x \to \pi} \frac{1+\cos x }{1+\cos 3x}=\frac{1}{9}$$
It was easy enough using L'Hôpital's rule, but I want to solve it without L'Hôpital.
I've tried manipulating the denominator with few simple trig identitities but nothing got me further than $\frac{1+\cos x }{1+\cos 3x}=\frac{1+\cos x}{\cos x(2\cos 2x-1)}$.
So I went to good all Wolfram Alpha which told me $$\frac{1+\cos x }{1+\cos 3x}=\frac{1}{(1-2\cos x)^2}$$ but I didn't get there, so any ideas on showing the equality above will be great.
I really prefer a clue or the first step as I believe I can do it myself- there's just something missing.
Thanks.

Answer 1

Enforcing $t=π-x$ and given the classical limit, $$\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$$ we have $$\lim_{t \to \pi} \frac{1+\cos t }{1+\cos 3t}\\=\lim_{x\to0}\frac{1-\cos x}{1-\cos 3x}=\frac19\lim_{x\to0}\frac{1-\cos x}{x^2}.\frac{(3x)^2}{1-\cos 3x} \\=\frac19\lim_{x\to0}\frac{1-\cos x}{x^2}.\lim_{x\to0}\frac{(3x)^2}{1-\cos 3x}\\=\frac19 \lim_{x\to0}\frac{1-\cos x}{x^2}.\lim_{h\to0}\frac{h^2}{1-\cos h}\\=\frac19$$

Answer 2

As $\cos(x+\pi)=-\cos x$ then it's the same as $$\lim_{x\to0}\frac{1-\cos x}{1-\cos 3x}.$$ As $$\cos3x=4\cos^3x-3\cos x,$$ it's the same as $$\lim_{x\to0}\frac{1-\cos x}{1+3\cos x-4\cos^3x} =\lim_{x\to0}\frac1{1+4\cos x+4\cos^2x}$$ etc.

Answer 3

I think that follows from straight multiplication

$(1+\cos(x))\times (1+4\cos^2(x)-4\cos(x))= 1+\cos(x)+4\cos^3(x)+4\cos^2(x)-4\cos(x)-4\cos^2(x)= 1+4\cos^3(x)-3\cos(x)=1+\cos(3x)$ which proves $$\frac{1+\cos(x)}{1+\cos(3x)}=\frac1{(1-2\cos(x))^2}$$

Answer 4

I always wonder why series expansions are often overlooked when calculating limits:

As shown above we need only consider for $x \rightarrow 0$ the following expression: $$\frac{1-\cos x}{1-\cos 3x}= \frac{1-(1-\frac{1}{2!}x^2 + \frac{1}{4!}x^4 \mp \cdots) }{1-(1-\frac{1}{2!}(3x)^2 + \frac{1}{4!}(3x)^4 \mp \cdots) } = \frac{\frac{1}{2!} - \frac{1}{4!}x^2 \pm \cdots }{\frac{9}{2!} - \frac{1}{4!}(3x)^2 \pm \cdots } \stackrel{x\rightarrow 0}{\longrightarrow}\frac{1}{9}$$