0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You can differentiate both sides, and for $x=0$ the two functions are equal. $\endgroup$ – Atmos Feb 10 '18 at 16:48
  • $\begingroup$ thx for the insight, we're not allowed using derivatives yet in this course though. $\endgroup$ – user114138 Feb 10 '18 at 16:52
0
$\begingroup$

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

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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: $$\frac{1-\cos x}{1-\cos 3x}= \frac{1-(1-\frac{1}{2!}x^2 + \frac{1}{4!}x^4 \mp \cdots) }{1-(1-\frac{1}{2!}(3x)^2 + \frac{1}{4!}(3x)^4 \mp \cdots) } = \frac{\frac{1}{2!} - \frac{1}{4!}x^2 \pm \cdots }{\frac{9}{2!} - \frac{1}{4!}(3x)^2 \pm \cdots } \stackrel{x\rightarrow 0}{\longrightarrow}\frac{1}{9}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.