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anonymous

  • one year ago

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→0 e^2x − e^−2x − 4x/x-sinx I've tried to do it w the rule but i keep getting it wrong, please help!!

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  1. caozeyuan
    • one year ago
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    \[\lim_{x \rightarrow 0}e ^{2x}-e ^{-2x}-\frac{ 4x }{ x }-\sin(x)\]

  2. caozeyuan
    • one year ago
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    am I correct in transcribing your question?

  3. anonymous
    • one year ago
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    almost, the whole thing is being divided by x-sinx

  4. anonymous
    • one year ago
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    For starters, is using the rule appropriate?

  5. anonymous
    • one year ago
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    well i plugged 0 in and i ended up with 0/0

  6. caozeyuan
    • one year ago
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    @SithsAndGiggles I learned this rule in calc 1 so I guess yes?

  7. caozeyuan
    • one year ago
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    @gaba a joke first: does you name means the neurotransmitter gamma-amino-butyic acid?

  8. caozeyuan
    • one year ago
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    I see what you mean so the correct version would be \[\lim_{x \rightarrow 0}\frac{ e^{2x}-e ^{-2x}-4x }{ x-\sin(x) }\]

  9. anonymous
    • one year ago
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    gaba its one of my nicknames, but yeah jaja

  10. anonymous
    • one year ago
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    yeah that's it

  11. caozeyuan
    • one year ago
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    Do you now what exactly is lhospital rule and how to use it?

  12. caozeyuan
    • one year ago
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    If you don't then that's the end of game, but if you do it's a piece of cake

  13. anonymous
    • one year ago
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    sort of, but i keep getting the same thing over and over again, so im confused

  14. caozeyuan
    • one year ago
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    Ok, so you know the new limit is formed by taking the quotient of the derivatives of the numerator and the denominator, right?

  15. anonymous
    • one year ago
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    yes

  16. caozeyuan
    • one year ago
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    so what are the derivatives we need?

  17. anonymous
    • one year ago
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    e^2x-e^-2x-4/x-cosx?

  18. caozeyuan
    • one year ago
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    OK, I have to say you kind of don't know your stuff

  19. caozeyuan
    • one year ago
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    \[\frac{ de ^{nx} }{ dx }= n*e^{nx}\]

  20. caozeyuan
    • one year ago
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    n is integer, for now

  21. caozeyuan
    • one year ago
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    you still there? what would be the correct answer?

  22. anonymous
    • one year ago
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    i got 2e^2x+2e^-2x-4 / 1-cosx

  23. caozeyuan
    • one year ago
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    but it is still 0 over 0 right?

  24. caozeyuan
    • one year ago
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    so you got to do one more round

  25. anonymous
    • one year ago
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    yeah, and i did but i kept getting that

  26. caozeyuan
    • one year ago
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    so 4e^2x+4e^2x/sin(x)

  27. caozeyuan
    • one year ago
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    wait.... that would be 8/0....WTF

  28. caozeyuan
    • one year ago
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    my brain is so rusty cuz I had my calc 1 3 years ago

  29. anonymous
    • one year ago
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    right now i have 16e^2x-16e^-2x / -sinx

  30. anonymous
    • one year ago
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    undefined right?

  31. caozeyuan
    • one year ago
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    how do you get your answer?! I didn't get any 16 or -sin

  32. anonymous
    • one year ago
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    What I meant by my comment was, does the limit satisfy the conditions for using L'Hopital's rule? The answer is yes, because substituting \(x=0\) directly gives you a proper indeterminate form: \[\frac{e^{2(0)}-e^{-2(0)}-4(0)}{0-\sin 0}=\frac{e^0-e^0-0}{0-0}=\frac{0}{0}\]

  33. anonymous
    • one year ago
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    i kept going, i had 2e^2x, and then 4e^2x, and then 8e^2x because i kept getting 0/0 with all of them

  34. caozeyuan
    • one year ago
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    But I stoped on the second trial cuz the upstair is 8

  35. anonymous
    • one year ago
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    \[\begin{align*}\lim_{x\to0}\frac{e^{2x}-e^{-2x}-4x}{x-\sin x}&=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[e^{2x}-e^{-2x}-4x\right]}{\dfrac{d}{dx}\left[x-\sin x\right]}\\\\ &=\lim_{x\to0}\frac{2e^{2x}+2e^{-2x}-4}{1-\cos x}=\frac{2e^0+2e^0-4}{1-\cos0}=\frac{0}{0}\\\\ &=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[2e^{2x}+2e^{-2x}-4\right]}{\dfrac{d}{dx}\left[1-\cos x\right]}\\\\ &=\lim_{x\to0}\frac{4e^{2x}-4e^{-2x}}{\sin x}=\frac{4e^0-4e^0}{\sin0}=\frac{0}{0}\\\\ &=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[4e^{2x}-4e^{-2x}\right]}{\dfrac{d}{dx}[\sin x]}\\\\ &=\lim_{x\to0}\frac{8e^{2x}+8e^{-2x}}{-\cos x}=\frac{8e^0+8e^0}{\cos0}\neq\frac{0}{0}\end{align*}\]

  36. anonymous
    • one year ago
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    so then my answer is -16?

  37. anonymous
    • one year ago
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    Sorry, I meant, not negative.

  38. anonymous
    • one year ago
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    Yes, just to check: http://www.wolframalpha.com/input/?i=Limit%5B%282+Sinh%5B2x%5D-4x%29%2F%28x-Sin%5Bx%5D%29%2Cx-%3E0%5D

  39. anonymous
    • one year ago
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    thank you so much!!

  40. anonymous
    • one year ago
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    yw

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