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MarcLeclair

  • 2 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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  1. MarcLeclair
    • 2 years ago
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    @zepdrix if you don't mind helping me with this one xD

  2. zepdrix
    • 2 years ago
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    \[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\]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,\[\huge \frac{0}{0}, \qquad \frac{\infty}{\infty}\]

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

  4. zepdrix
    • 2 years ago
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    Woops, I wrote that incorrectly :) one sec.

  5. zepdrix
    • 2 years ago
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    \[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\quad=\quad \huge \lim_{x \rightarrow 0^+} \frac{2\cdot \ln x}{\csc x}\]

  6. MarcLeclair
    • 2 years ago
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    thats infinity / 0 . we're still not allowed to apply that?

  7. MarcLeclair
    • 2 years ago
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    I mean L'hospital rule*

  8. zepdrix
    • 2 years ago
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    Is it? Hmm lemme think.

  9. MarcLeclair
    • 2 years ago
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    oh wait no thats infinity

  10. zepdrix
    • 2 years ago
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    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.

  11. MarcLeclair
    • 2 years ago
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    That isn't the answer ahaha. I hate cal 1 :(

  12. zepdrix
    • 2 years ago
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    That's not the right path to take? :o

  13. zepdrix
    • 2 years ago
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    Seems like we're on the right track :)

  14. MarcLeclair
    • 2 years ago
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    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

  15. MarcLeclair
    • 2 years ago
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    * I can verify the answer but it doesn't give it per say

  16. zepdrix
    • 2 years ago
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    No we're not done yet :) lol

  17. zepdrix
    • 2 years ago
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    We have successfully gotten it into the form infty / infty, from here we're allowed to apply L'Hop, yes?

  18. MarcLeclair
    • 2 years ago
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    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?

  19. MarcLeclair
    • 2 years ago
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    err i mean -csctanx

  20. zepdrix
    • 2 years ago
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    -cscx cotx I think

  21. MarcLeclair
    • 2 years ago
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    yup -.- so I meant 2(1/x)/-cscxcotx

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

  23. MarcLeclair
    • 2 years ago
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    yup and replace sin with 1-cos(square)x sorry french keyboard :P

  24. zepdrix
    • 2 years ago
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    Hmm it looks like it might still be giving us trouble :( Hmmmm

  25. MarcLeclair
    • 2 years ago
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    second derivative?

  26. KennethR
    • 2 years ago
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    Its now in a 0/0 form which means that you can apply L'Hop again

  27. zepdrix
    • 2 years ago
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    But won't it just get worse..? :d

  28. phi
    • 2 years ago
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    you could use lim x->0 sin(x)/x = 1

  29. zepdrix
    • 2 years ago
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    Oh derp :o good call.

  30. phi
    • 2 years ago
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    leaving -2 tan(x) = 0

  31. MarcLeclair
    • 2 years ago
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    and phil is right xD it is so hard to see through all that. I wish i could best response both :P

  32. zepdrix
    • 2 years ago
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    \[\large \lim_{x \rightarrow 0^+}\frac{-2 \cdot \sin^2x}{x \cos x} \quad = \quad -2\lim_{x \rightarrow 0^+}\frac{\sin x}{x} \cdot \lim_{x \rightarrow 0^+} \frac{\sin x}{\cos x}\]

  33. MarcLeclair
    • 2 years ago
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    yeah I didn't see that :/

  34. zepdrix
    • 2 years ago
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    Yah that's neato :O that limit slipped my mind heh

  35. KennethR
    • 2 years ago
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    Ah, thats clever

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