# Thread: Trigonometric and logarithmic limit?

1. ## Trigonometric and logarithmic limit?

Evaluate:
$\lim_{x\rightarrow \infty} x^2\sin\left(\log \sqrt{\cos \frac{\pi}{x}}\right)$

Spoiler:
$-\frac{\pi^2}{4}$

I am unable to get the required answer. How to do it?

2. $\lim_{x\rightarrow \infty} x^2\sin\left(\log \sqrt{\cos \frac{\pi}{x}}\right)$

$
=\lim_{x\rightarrow \infty} x^2\frac{\sin\left(\log \sqrt{\cos \frac{\pi}{x}}\right)}{log\sqrt{\cos\frac{\pi}{x}} }
$
$log\sqrt{\cos\frac{\pi}{x}}$

$=-\pi^2\lim_{x\rightarrow \infty} \frac{\sin\left(\log \sqrt{\cos \frac{\pi}{x}}\right)}{log\sqrt{\cos\frac{\pi}{x}} }
$
. $\frac{log(1+\sqrt{\cos\frac{\pi}{x}}-1)}{\sqrt{cos\frac{\pi}{x}}-1}$. $\frac{1-\sqrt{cos\frac{\pi}{x}}}{\frac{\pi^2}{x^2}}$

$=-\pi^2\lim_{x\rightarrow \infty} \frac{\sin\left(\log \sqrt{\cos \frac{\pi}{x}}\right)}{log\sqrt{\cos\frac{\pi}{x}} }
$
. $\frac{log(1+\sqrt{\cos\frac{\pi}{x}}-1)}{\sqrt{cos\frac{\pi}{x}}-1}$. $\frac{1-\cos\frac{\pi}{x}}{\frac{\pi^2}{x^2}}$. $\frac{1}{1+\cos\frac{\pi}{x}}$

$
=-\frac{\pi^2}{4}
$

3. A little explanation of the last part:

$\lim_{x\to \infty}\frac{1-\cos\frac{\pi}{x}}{\frac{\pi^2}{x^2}}=\lim_{t\to 0}\frac{1-\cos t}{t^2}=\lim_{t\to 0}\frac{1-\cos t}{t^2}\frac{1+\cos t}{1+\cos t}$ $=\lim_{t\to 0}\frac{\sin^2 t}{t^2(1+\cos t)}=\lim_{t\to 0}\frac{\sin t}{t}\frac{\sin t}{t}\frac{1}{1+\cos t}=\frac{1}{2}$

4. Does $\lim_{x\rightarrow \infty} \frac{log(1+\sqrt{\cos\frac{\pi}{x}}-1)}{\sqrt{cos\frac{\pi}{x}}-1} = 1$?

EDIT: I take that back. It's indeterminate in that form. What am I missing?

5. Make the substitution $t=\sqrt{\cos {\frac{\pi}{x}}}-1$ and you get $t\to 0$ as $x\to \infty.$

Standard limit: $\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$