Find the limit of
$$\lim_{x\to0}\left[1 + \left(\frac{\log \cos x}{\log \cos(x/2)}\right)^2 \right]^2$$
Any help would be thoroughly appreciated.
Asked
Active
Viewed 2,878 times
8
-
I've edited your question with MathJax, please use mathjax in the future. Note that $\log_b a=\frac{\log a}{\log b}$ – Alice Ryhl Oct 23 '14 at 13:43
-
1I have a doubt that the limit is even well-defined, since there are infinitely many $x$ such that $\cos x<0$ or $\cos(x/2)<0$. – Hanul Jeon Oct 23 '14 at 13:47
-
1Sorry the limit is to zero – user34304 Oct 23 '14 at 13:48
-
The answers 289 – user34304 Oct 23 '14 at 13:49
-
I guess x is an element of $]-\frac{\pi}{2}, \frac{\pi}{2}[$, in which case the cos is positive and x can ->0 without pb – mvggz Oct 23 '14 at 13:52
-
As shown in comments and answer, just do it slowly looking at the pieces. Cheers :-) – Claude Leibovici Oct 23 '14 at 14:03
5 Answers
6
$$\lim_{x\to0}\left[1 + \left(\frac{\log \cos x}{\log \cos(x/2)}\right)^2 \right]^2$$ $$\left[1+\lim_{x\to0}\left(\frac{\log \cos x}{\log \cos(x/2)}\right)^2 \right]^2$$ $$\left[1+\left(\lim_{x\to0}\frac{\log \cos x}{\log \cos(x/2)}\right)^2 \right]^2$$ Apply L'Hopital $$\left[1+\left(2\cdot\lim_{x\to0}\frac{\tan x}{\tan(x/2)}\right)^2 \right]^2$$ Apply L'Hopital $$\left[1+\left(4\cdot\lim_{x\to0}\frac{\sec^2 x}{\sec^2(x/2)}\right)^2 \right]^2$$ $$\left[1+\left(4\cdot\frac{\sec^20}{\sec^20}\right)^2 \right]^2$$ $$\left[1+\left(4\cdot\frac{1}{1}\right)^2 \right]^2=289$$
Alice Ryhl
- 7,853
-
-
1@user34304 $$\lim_{x\to a} f(g(x))=f\left(\lim_{x\to a}g(x)\right)$$ if $f(x)$ is continuous. – Alice Ryhl Oct 23 '14 at 14:14
4
Hint: use the fact that $$ \log x \sim_{x\to 1} x-1\\ \cos u =_{u\to 0} 1 - \frac{u^2}2 + o(u^2) $$
then you get $$ \log \cos x \sim-\frac{x^2}2\\ \log \cos \frac x2 \sim-\frac{x^2}8\\ \frac{\log \cos x}{\log \cos \frac x2} \to 4 $$
so the final limit is $$ (1 + 4^2)^2 = 17^2 = 289 $$
mookid
- 28,236
0
From the Taylor formula $\log(\cos x)$ behaves as $x^2/2$ near 0, hence the limit is $(1+4^2)^2=17^2$.
Przemysław Scherwentke
- 13,668
- 5
- 35
- 56
0
$t=\cos(x/2)\to 1$
$\frac{\log \cos x}{\log \cos(x/2)}=\frac{\log 2\cos^2 (\frac{x}{2})-1}{\log \cos(x/2)}=\frac{\log 2t^2-1}{\log t}$
$\lim_{x\to 0}\frac{\log \cos x}{\log \cos(x/2)}=\lim_{t\to 1}\frac{\log (2t^2-1)}{\log t}=4$ by L'Hospital's rule.
Then use composite rule of limit.
John
- 13,204
