Help with this L'hospital rule problem:
Solve for the following limit:
Lim as x approaches 0(+) of 2sinxlnx
I don't know what to do, I'm always stuck with a ln0 which is undefined no?

- anonymous

- katieb

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- anonymous

@zepdrix if you don't mind helping me with this one xD

- zepdrix

\[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\]As x approaches 0 from the right, the limit is approaching,\[\huge 2\cdot 0\cdot -\infty\]
This is of indeterminate form.
But it's not one of the forms we're allowed to apply L'Hop to.
Remember we need one of these two forms,\[\huge \frac{0}{0}, \qquad \frac{\infty}{\infty}\]

- zepdrix

We'll have to apply some fancy trig work.\[\large \sin x=\frac{1}{\csc x}\]Substituting gives us,\[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\quad=\quad \huge \lim_{x \rightarrow 0^+} \frac{2\cdot \ln x}{\frac{1}{\csc x}}\]

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## More answers

- zepdrix

Woops, I wrote that incorrectly :) one sec.

- zepdrix

\[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\quad=\quad \huge \lim_{x \rightarrow 0^+} \frac{2\cdot \ln x}{\csc x}\]

- anonymous

thats infinity / 0 . we're still not allowed to apply that?

- anonymous

I mean L'hospital rule*

- zepdrix

Is it? Hmm lemme think.

- anonymous

oh wait no thats infinity

- zepdrix

csc x = 1/sin x, as sin x gets closer to 0, the fraction gets bigger and bigger,
Yah I think you're right, infinity.

- anonymous

That isn't the answer ahaha. I hate cal 1 :(

- zepdrix

That's not the right path to take? :o

- zepdrix

Seems like we're on the right track :)

- anonymous

I don't know I just have the question, no answer, I dwelled on it for a good hour nothing came to mind. I just know its not infinity :P

- anonymous

* I can verify the answer but it doesn't give it per say

- zepdrix

No we're not done yet :) lol

- zepdrix

We have successfully gotten it into the form infty / infty, from here we're allowed to apply L'Hop, yes?

- anonymous

yup so I got 2 1/x / csctanx which is 1/sinx * sinx/cos x which gives me 2(1/x)/(1/cosx) am i right?

- anonymous

err i mean -csctanx

- zepdrix

-cscx cotx I think

- anonymous

yup -.- so I meant 2(1/x)/-cscxcotx

- zepdrix

Which... after you move things around... I think we get,\[\huge \lim_{x \rightarrow 0^+}\frac{-2 \cdot \sin^2x}{x \cos x}\]

- anonymous

yup and replace sin with 1-cos(square)x sorry french keyboard :P

- zepdrix

Hmm it looks like it might still be giving us trouble :( Hmmmm

- anonymous

second derivative?

- anonymous

Its now in a 0/0 form which means that you can apply L'Hop again

- zepdrix

But won't it just get worse..? :d

- phi

you could use lim x->0 sin(x)/x = 1

- zepdrix

Oh derp :o good call.

- phi

leaving -2 tan(x) = 0

- anonymous

and phil is right xD it is so hard to see through all that. I wish i could best response both :P

- zepdrix

\[\large \lim_{x \rightarrow 0^+}\frac{-2 \cdot \sin^2x}{x \cos x} \quad = \quad -2\lim_{x \rightarrow 0^+}\frac{\sin x}{x} \cdot \lim_{x \rightarrow 0^+} \frac{\sin x}{\cos x}\]

- anonymous

yeah I didn't see that :/

- zepdrix

Yah that's neato :O that limit slipped my mind heh

- anonymous

Ah, thats clever

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