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Math Help - Evaluate limits?

  1. #1
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    Evaluate limits?

    Evaluate the following limits:

    1) lim as x-->0+, x^(1/3)*ln(x)

    2) lim as x-->1+, (1/ln(x) - 1/(x-1))

    3) lim as x-->infinity, (1+(4/x))^(x/5)

    I'm unsure how to even begin these ones...any help would be greatly appreciated.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by demonmath View Post
    Evaluate the following limits:

    1) lim as x-->0+, x^(1/3)*ln(x)

    2) lim as x-->1+, (1/ln(x) - 1/(x-1))

    3) lim as x-->infinity, (1+(4/x))^(x/5)

    I'm unsure how to even begin these ones...any help would be greatly appreciated.
    Do you know your indeterminate forms?
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  3. #3
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    I'm going to say no...as i don't know for sure what that means.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by demonmath View Post
    I'm going to say no...as i don't know for sure what that means.
    It would help me if I knew if you were in Calc 1 or Calc 2. Are you supposed to be applying L'Hopital's Rule?
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  5. #5
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    I'm in Calc 251, I'm not sure what the equivalent of that is. No instructions were given on the practice assignment question I posted except what I've included. There were four questions and I answered the first one using L'Hopitals, so I'm assuming there would be no problem with this.
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by demonmath View Post
    I'm in Calc 251, I'm not sure what the equivalent of that is. No instructions were given on the practice assignment question I posted except what I've included. There were four questions and I answered the first one using L'Hopitals, so I'm assuming there would be no problem with this.
    Ok. So, you instructer shouldv'e mentioned something about indeterminate forms, but anyway.

    Not that by substitution, number 2 yields the indeterminate form \infty-\infty. The goal here is to rewrite this into a function that produces - by substitution - either \frac{0}{0} or \frac{\infty}{\infty}, so that L'Hopital's rule will apply. Can you do this?
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    i think i did...i just multiply to get a common denominator right?
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by demonmath View Post
    i think i did...i just multiply to get a common denominator right?
    Indeed! then apply L'Hopital! See the magic happen...
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  9. #9
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    i can get the first one now from your help as well but im struggling with the last one still. If you could lend a hand again that would be great!
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by demonmath View Post
    i can get the first one now from your help as well but im struggling with the last one still. If you could lend a hand again that would be great!
    Sure.

    Try letting y=\lim_{x\to\infty}\left(1+\frac{4}{x}\right)^{x/5}

    Now take the ln of both sides to get that nasty exponent out if there..
    Last edited by mr fantastic; November 2nd 2009 at 07:44 PM. Reason: Fixed a typo in the exponent: x/5 not 5/x
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  11. #11
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    okay, so the exponent comes out front then. So should i get on the right side:

    [ln(1+4/x)]/(5x^-1)

    Or is there a better way to do it
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  12. #12
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    Quote Originally Posted by VonNemo19 View Post
    Sure.

    Try letting y=\lim_{x\to\infty}\left(1+\frac{4}{x}\right)^{x/5}

    Now take the ln of both sides to get that nasty exponent out if there..
    A well known limit is \lim_{u \to +\infty}\left(1+\frac{a}{u}\right)^{u} = e^a.

    So the asked for limit is obvious if you make the substitution u = \frac{x}{5}: \lim_{u\to +\infty}\left(1+\frac{4/5}{u}\right)^{u}.
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  13. #13
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    Not as obvious as I would have hoped, clearly. Im still stuck.
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  14. #14
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    Quote Originally Posted by demonmath View Post
    Not as obvious as I would have hoped, clearly. Im still stuck.
    Look at the well known limit I pointed out to you. Look at what your limit becomes after the substitution is made. Compare the two. Identify the value of a. Write down the value of your limit.
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