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Homework Help: Limits with e^x and an integral too!

  1. Feb 7, 2006 #1

    G01

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    Hi I'm having some trouble with evaluating these limits. I can't figure out what to do. I guess i forgot some calc one. I don't have much work but All I'm asking for for now is a couple hints.

    [tex]\lim_{x\rightarrow\infty} \frac{e^{3x} -e^{-3x}}{e^{3x} + e^{-3x}} [/tex]

    I tried dividing numerator and denominator by e^3x and 3^-3x Neither worked? Any hints on what else to do?

    [tex] \lim_{x\rightarrow2^+} e^{3/(2-x)} [/tex] I have no idea here. Please I know theres not much work but Im totaly lost.

    Now the integral:

    [tex] \int e^x \sqrt{1+e^x} dx [/tex]

    I tried the substitution u = [tex] \sqrt{1+e^x} [/tex]

    Using that substitution i get this:

    [tex] 2\int u^2 du [/tex]

    integrate this and you get:

    [tex] (1/2) u^4 + C [/tex]

    which is:

    (1+ 2e^x +e^2x)/2, This is nowhere near the answer. Where did I go wrong. I don't see why that substitution didn't work.
     
    Last edited: Feb 7, 2006
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  3. Feb 7, 2006 #2

    G01

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    wait i saw my mistake on the integral. Nevermind there.
     
  4. Feb 7, 2006 #3
    Hello. I'll offer some quick assistance with the integral

    [tex] \int e^x \sqrt{1+e^x} dx [/tex]

    For this, try setting [tex] u = [/tex] [tex] 1 + e^x [/tex]
    Thus,
    [tex] du = e^x \cdot dx [/tex]

    Does this help?
     
  5. Feb 7, 2006 #4

    krab

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    try it in a calculator; it converges very quickly. This is because the second term on top and second term on bottom very quickly become utterly negligible.
     
  6. Feb 7, 2006 #5

    G01

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    Krab and ebola thank you. I solved the integral and the first limit. Now if Anybody can help on the second limit, thatd be great.
     
  7. Feb 7, 2006 #6
    [tex] \lim_{x\rightarrow2^+} e^{3/(2-x)} [/tex]

    Well as x approaches two, the denominator becomes a very small positive number, let's call it A, and 3/A becomes inifinitely large. Once this infinitely growing number is used as an exponent for e, I would say the limit is positive infinity.
     
  8. Feb 7, 2006 #7

    G01

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    That not the answer. The answer in the back is 0. I can't figure out how to manipulate that so I can get that answer
     
  9. Feb 8, 2006 #8

    benorin

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    [tex] \lim_{x\rightarrow2^+} e^{3/(2-x)} = e^{\lim_{x\rightarrow2^+} 3/(2-x)}[/tex] by continuity of the exponential function and [tex]\frac{3}{2-x} \rightarrow -\infty \mbox{ as }x\rightarrow2^+[/tex] so one may put

    [tex]\lim_{x\rightarrow2^+} e^{3/(2-x)} = \lim_{u\rightarrow +\infty} e^{u}=0 [/tex]
     
  10. Feb 8, 2006 #9

    benorin

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    tanh(3x) limit at infinity

    [tex]\lim_{x\rightarrow\infty} \frac{e^{3x} -e^{-3x}}{e^{3x} + e^{-3x}} =\lim_{x\rightarrow\infty} \frac{e^{3x} -\frac{1}{e^{3x}}}{e^{3x} + \frac{1}{e^{3x}}}=\lim_{x\rightarrow\infty} \frac{\frac{e^{6x}-1}{e^{3x}}}{\frac{e^{6x}+1}{e^{3x}}}=\lim_{x\rightarrow\infty} \frac{e^{6x}-1}{e^{6x}+1}[/tex]
    [tex]=\lim_{x\rightarrow\infty} \frac{e^{6x}-1}{e^{6x}+1}\cdot\frac{e^{-6x}}{e^{-6x}}= \lim_{x\rightarrow\infty} \frac{1-e^{-6x}}{1+e^{-6x}}=\frac{1-0}{1+0}=1[/tex]
     
  11. Feb 8, 2006 #10

    VietDao29

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    Whoops, typo here, benorin.
    If u tends to positive infinity, then it should be:
    [tex]\lim_{u\rightarrow +\infty} e^{u}= + \infty[/tex]
    In this problem, u tends to negative infinity, so it should be:
    [tex]\lim_{u\rightarrow -\infty} e^{u}= 0[/tex] (it's negative infinity, not positive infinity).
    By the way, may I suggest you not to post a COMPLETE solution.
    If I recalled correctly, this is the third time I catch you posting a COMPLETE solution. :)
     
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