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Homework Help: Limit involving L'Hopital's Rule

  1. Sep 19, 2012 #1
    1. The problem statement, all variables and given/known data

    \displaystyle\lim_{x\rightarrow 0^+} \frac{e^x - (1 + x)}{x^n}

    where n is a positive integer

    3. The attempt at a solution

    The numerator approaches 0 as x approaches 0 from the right. The denominator approaches 0 with whichever positive integer, n. This gives an indeterminate form, so I can apply L'Hopital's Rule:

    \displaystyle\lim_{x\rightarrow 0^+} \frac{e^x - 1}{nx^{n-1}}

    Now, here's where I'm not too sure...

    The numerator will still approach 0. The denominator will still approach 0 as x approaches 0, it's just being multiplied by some constant. If I apply L'Hopital's Rule again:

    \displaystyle\lim_{x\rightarrow 0^+} \frac{e^x}{n(n-1)x^{n-2}}

    The numerator now reaches 1 as x approaches 0.

    I'm unsure of how to evaluate the limit because I'm unsure of how to treat the denominator.

    If n was some large positive integer, I'd just get (1/0) as x approaches 0 which would yield to infinity... But if n had been some very small integer, the denominator would have eventually simplified to some constant term and evaluating the limit would not give infinity.

    How do I actually go on to evaluate the limit without assigning a value for n?

    Would it be legal to just specify the two specific cases (n = 1 and n = 2) where the limit doesn't evaluate towards infinity and then just give the general case for n > 2?
    Last edited: Sep 19, 2012
  2. jcsd
  3. Sep 19, 2012 #2
    Do you know the series expansion for ex? Makes it a lot easier.

    (But yes, you would consider each case)
  4. Sep 19, 2012 #3
    Nope. Don't know the series expansion for anything... Don't know what that is.

    I'm in the first couple weeks of Calc 2.

    I just wanted to make sure my thought process was correct throughout. Thanks.
  5. Sep 19, 2012 #4


    Staff: Mentor

    After two applications of L'Hopital's Rule, the numerator is approaching 1 and the denominator is approaching 0 (but remains positive), so the expression is approaching infinity.
  6. Sep 19, 2012 #5


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    And, as long as n> 1, the denominator goes to 0. What does that tell you?

    Do the cases n= 0 and n= 1 separately.

  7. Sep 19, 2012 #6
    I'm a little confused now...

    Wouldn't n = 0 be omitted since there's a restriction based on n being only positive integers?

    From what I understand, n = 1 and n = 2 are the two cases where the expression doesn't grow infinitely large. Any integer larger than 2 will leave (1/0). Shouldn't it be n > 2?
  8. Sep 19, 2012 #7


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    It looks to me that you are correct on all accounts.

    So, what do you get for each case: n = 1 and n = 2 ?
  9. Sep 19, 2012 #8
    I worked it out and got at n = 1, the limit is 0 and at n = 2, the limit is 1/2
  10. Sep 19, 2012 #9


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    Excellent !
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