indeterminate form

**xxlvh** · November 4th 2009, 08:55 PM

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.

**tonio** · November 4th 2009, 09:00 PM

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

**xxlvh** · November 5th 2009, 04:21 PM

Sorry, I'm still puzzled on how to "separate" them?

**Krizalid** · November 5th 2009, 05:18 PM

$\sin (x)\ln (x)=\frac{\sin x}{x}\cdot x\ln x.$

**tonio** · November 6th 2009, 02:10 AM

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

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