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MarcLeclair
Group Title
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?
 one year ago
 one year ago
MarcLeclair Group Title
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?
 one year ago
 one year ago

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MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix if you don't mind helping me with this one xD
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\]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,\[\huge \frac{0}{0}, \qquad \frac{\infty}{\infty}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
We'll have to apply some fancy trig work.\[\large \sin x=\frac{1}{\csc x}\]Substituting gives us,\[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\quad=\quad \huge \lim_{x \rightarrow 0^+} \frac{2\cdot \ln x}{\frac{1}{\csc x}}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Woops, I wrote that incorrectly :) one sec.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\quad=\quad \huge \lim_{x \rightarrow 0^+} \frac{2\cdot \ln x}{\csc x}\]
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
thats infinity / 0 . we're still not allowed to apply that?
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
I mean L'hospital rule*
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Is it? Hmm lemme think.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
oh wait no thats infinity
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
That isn't the answer ahaha. I hate cal 1 :(
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
That's not the right path to take? :o
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Seems like we're on the right track :)
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
* I can verify the answer but it doesn't give it per say
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
No we're not done yet :) lol
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
We have successfully gotten it into the form infty / infty, from here we're allowed to apply L'Hop, yes?
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
err i mean csctanx
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
cscx cotx I think
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
yup . so I meant 2(1/x)/cscxcotx
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Which... after you move things around... I think we get,\[\huge \lim_{x \rightarrow 0^+}\frac{2 \cdot \sin^2x}{x \cos x}\]
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
yup and replace sin with 1cos(square)x sorry french keyboard :P
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Hmm it looks like it might still be giving us trouble :( Hmmmm
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
second derivative?
 one year ago

KennethR Group TitleBest ResponseYou've already chosen the best response.0
Its now in a 0/0 form which means that you can apply L'Hop again
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
But won't it just get worse..? :d
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
you could use lim x>0 sin(x)/x = 1
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Oh derp :o good call.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
leaving 2 tan(x) = 0
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
and phil is right xD it is so hard to see through all that. I wish i could best response both :P
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large \lim_{x \rightarrow 0^+}\frac{2 \cdot \sin^2x}{x \cos x} \quad = \quad 2\lim_{x \rightarrow 0^+}\frac{\sin x}{x} \cdot \lim_{x \rightarrow 0^+} \frac{\sin x}{\cos x}\]
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
yeah I didn't see that :/
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Yah that's neato :O that limit slipped my mind heh
 one year ago

KennethR Group TitleBest ResponseYou've already chosen the best response.0
Ah, thats clever
 one year ago
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