L'Hospital problem

**stiitches** · November 4th 2010, 08:41 PM

Another one:

this time im lost

**mr fantastic** · November 4th 2010, 09:42 PM

Originally Posted by stiitches

Another one:

this time im lost

From your post title you are obviously aware of how it's to be done. So what have you tried and where are you stuck?

**pirateboy** · November 4th 2010, 09:50 PM

Recall, L'Hospital's rule states that if

$\displaystyle \lim_{x\to n}\frac{f(x)}{g(x)} = \frac{0}{0} \;\;\text{or}\;\; \frac{\pm \infty}{\pm \infty}\quad \text{then}\quad \lim_{x\to n} \frac{f(x)}{g(x)} \overset{\text{H}}{=}\lim_{x\to n}\frac{f'(x)}{g'(x)}$

So....

$\displaystyle \lim_{x\to 1}\frac{\ln(x)}{\sin(2\pi x)}\overset{\text{H}}{=}\lim_{x\to 1}\frac{\frac{d}{dx}\ln(x)}{\frac{d}{dx}\sin(2\pi x)}$

I'm confident you can handle it from here.

**pirateboy** · November 4th 2010, 10:02 PM

oops! i left our a few conditions. $f\;\text{and}\;g$ must be differentiable and $g'(x) \neq 0$ .

But both are true here. So it shouldn't matter. Just bust out some sweet differentiation moves and you're golden.

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