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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?

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@zepdrix if you don't mind helping me with this one xD
\[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\]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,\[\huge \frac{0}{0}, \qquad \frac{\infty}{\infty}\]
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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Other answers:

Woops, I wrote that incorrectly :) one sec.
\[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\quad=\quad \huge \lim_{x \rightarrow 0^+} \frac{2\cdot \ln x}{\csc x}\]
thats infinity / 0 . we're still not allowed to apply that?
I mean L'hospital rule*
Is it? Hmm lemme think.
oh wait no thats infinity
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.
That isn't the answer ahaha. I hate cal 1 :(
That's not the right path to take? :o
Seems like we're on the right track :)
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
* I can verify the answer but it doesn't give it per say
No we're not done yet :) lol
We have successfully gotten it into the form infty / infty, from here we're allowed to apply L'Hop, yes?
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?
err i mean -csctanx
-cscx cotx I think
yup -.- so I meant 2(1/x)/-cscxcotx
Which... after you move things around... I think we get,\[\huge \lim_{x \rightarrow 0^+}\frac{-2 \cdot \sin^2x}{x \cos x}\]
yup and replace sin with 1-cos(square)x sorry french keyboard :P
Hmm it looks like it might still be giving us trouble :( Hmmmm
second derivative?
Its now in a 0/0 form which means that you can apply L'Hop again
But won't it just get worse..? :d
  • phi
you could use lim x->0 sin(x)/x = 1
Oh derp :o good call.
  • phi
leaving -2 tan(x) = 0
and phil is right xD it is so hard to see through all that. I wish i could best response both :P
\[\large \lim_{x \rightarrow 0^+}\frac{-2 \cdot \sin^2x}{x \cos x} \quad = \quad -2\lim_{x \rightarrow 0^+}\frac{\sin x}{x} \cdot \lim_{x \rightarrow 0^+} \frac{\sin x}{\cos x}\]
yeah I didn't see that :/
Yah that's neato :O that limit slipped my mind heh
Ah, thats clever

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