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anonymous
 4 years ago
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
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@zepdrix if you don't mind helping me with this one xD

zepdrix
 4 years ago
Best 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,\[\huge \frac{0}{0}, \qquad \frac{\infty}{\infty}\]

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

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Woops, I wrote that incorrectly :) one sec.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats infinity / 0 . we're still not allowed to apply that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I mean L'hospital rule*

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Is it? Hmm lemme think.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh wait no thats infinity

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1csc 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
 4 years ago
Best ResponseYou've already chosen the best response.0That isn't the answer ahaha. I hate cal 1 :(

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1That's not the right path to take? :o

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Seems like we're on the right track :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0* I can verify the answer but it doesn't give it per say

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1No we're not done yet :) lol

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup 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
 4 years ago
Best ResponseYou've already chosen the best response.0yup . so I meant 2(1/x)/cscxcotx

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup and replace sin with 1cos(square)x sorry french keyboard :P

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Hmm it looks like it might still be giving us trouble :( Hmmmm

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Its now in a 0/0 form which means that you can apply L'Hop again

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1But won't it just get worse..? :d

phi
 4 years ago
Best ResponseYou've already chosen the best response.1you could use lim x>0 sin(x)/x = 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and phil is right xD it is so hard to see through all that. I wish i could best response both :P

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah I didn't see that :/

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Yah that's neato :O that limit slipped my mind heh
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