# indeterminate form

• November 4th 2009, 08:55 PM
xxlvh
indeterminate form
I'm having trouble figuring this one out:

$\lim_{x\to0^+}sin(x)ln(x)$

$= \lim_{x\to0^+}\frac{sin(x)}{\frac{1}{ln(x)}}$

$= \lim_{x\to0^+}\frac{cos(x)}{\frac{-1}{x(ln(x))^2}}$

Which doesn't seem to help very much.
• November 4th 2009, 09:00 PM
tonio
Quote:

Originally Posted by xxlvh
I'm having trouble figuring this one out:

$\lim_{x\to0^+}sin(x)ln(x)$

$= \lim_{x\to0^+}\frac{sin(x)}{\frac{1}{ln(x)}}$

$= \lim_{x\to0^+}\frac{cos(x)}{\frac{-1}{x(ln(x))^2}}$

Which doesn't seem to help very much.

Do it the other way: $\frac{\ln x}{\frac{1}{\sin x}}$ and then "separate" the result in two convenient factors.

The limit is zero.

Tonio
• November 5th 2009, 04:21 PM
xxlvh
Sorry, I'm still puzzled on how to "separate" them?
• November 5th 2009, 05:18 PM
Krizalid
$\sin (x)\ln (x)=\frac{\sin x}{x}\cdot x\ln x.$
• November 6th 2009, 02:10 AM
tonio
Quote:

Originally Posted by Krizalid
$\sin (x)\ln (x)=\frac{\sin x}{x}\cdot x\ln x.$

This, of course, is much better and simpler than what I proposed, which was:

$\lim_{x\to 0}\sin x\ln x =\lim_{x\to 0} \frac{\ln x}{\frac{1}{\sin x}}=\lim_{x\to 0}\frac{\frac{1}{x}}{\frac{-\cos x}{\sin^2x}}=\lim_{x\to 0}-\frac{\sin x}{x}\frac{\sin x}{\cos x}$ $=(-1)\cdot 0=0$

Tonio