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Thread: L'Hopital's Rule and Derivatives

  1. #1
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    L'Hopital's Rule and Derivatives

    I tried solving the following problem many times, but each time I do I end up with the wrong answer. Is the order which I differentiated this expression in correct? Also, is the derivative of e (a constant) e or 0? I'm still very new to calculus, so I'd really appreciate a detailed explanation in regarsds to solving this problem. Thanks.

    $\displaystyle \displaystyle 2(1-e^2){\lim_{n \to \infty}(\frac{\frac{e^{2/n}}{n}}{1 - e^{2/n}})$

    $\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{\frac{(n(e^{2/n})' - (e^{2/n})n')}{n^2}}{1' - (\frac{2}{n}(e^{2/n - 1})e')}$

    $\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{\frac{(n(\frac{2}{n}(e^{2/n - 1})e) - (e^{2/n})1)}{n^2}}{-(\frac{2}{n}(e^{2/n - 1})e)}$

    $\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{\frac{(2(e^{2/n}) - (e^{2/n})}{n^2}}{-(\frac{2}{n}(e^{2/n})}$

    $\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}(\frac{(2(e^{2/n}) - (e^{2/n})}{n^2})(\frac{1}{-{2}{n}(e^{2/n})})$

    $\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{2(e^{2/n}) - (e^{2/n})}{n^2(-\frac{2}{n}(e^{2/n}))}$

    $\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{2(e^{2/n}) - (e^{2/n})}{(-2n(e^{2/n}))}$

    $\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{(e^{2/n})(2 - 1)}{-2n(e^{2/n})}$

    $\displaystyle = \displaystyle 2(1-e^2){\lim_{n \to \infty}\frac{1}{-2n}$

    $\displaystyle = \displaystyle 2(1-e^2)(0)$

    $\displaystyle = 0 $
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  2. #2
    Senior Member yeKciM's Avatar
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    it would be much easier if u wrote it like


    $\displaystyle \displaystyle 2(1-e^2) \lim _{n \to \infty} \frac {e^{2/n}}{n(1-e^{2/n})}$

    and then apply it
    Last edited by yeKciM; Jul 31st 2010 at 10:42 AM.
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  3. #3
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    Simplify it first. Also, the derivative of any constant is 0. e is no exception.
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  4. #4
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    What and when do I simplify?
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  5. #5
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    The way yeKciM wrote is the simplified version I was referencing .
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  6. #6
    Senior Member yeKciM's Avatar
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    then just apply the :

    $\displaystyle \displaystyle (\frac{u}{v})' = \frac {u'v-v'u}{v^2} $

    theres little a bit or work to simplify it after derivation but just go easy
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  7. #7
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    Thanks. I'll try that right now.
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  8. #8
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    After applying the quotient rule, I ended up with:

    $\displaystyle \frac{((n(1 - e^{2/n}))(e^{2/n})' - (e^{2/n})(n(1 - e^{2/n}))'}{(n(1 - e^{2/n}))^2}$

    In order to simplify, is it necessary that I find the derivatives of the expressions accompanied by the prime $\displaystyle '$ notation? Or do I ignore the primes and begin by converting this one fraction into two of them?
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  9. #9
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    You are making things complicated for yourself.

    L'Hopital's rule is used when the limit of your right-hand factor $\displaystyle \displaystyle\huge\frac{\frac{e^{\frac{2}{n}}}{n}} {1-e^{\frac{2}{n}}}$

    approaches $\displaystyle \frac{0}{0}$ as n goes to infinity.

    It's already in that form, but the differentiation wrt n is inconvenient due to
    n being in a denominator position.

    Therefore substitute $\displaystyle x=\frac{1}{n}$ and evaluate the limit as x goes to zero instead.


    $\displaystyle \displaystyle\lim_{n\rightarrow\infty}\frac{\frac{ e^{\frac{2}{n}}}{n}}{1-e^{\frac{2}{n}}}$

    $\displaystyle =\displaystyle\lim_{x\rightarrow\ 0}\frac{xe^{2x}}{1-e^{2x}}$

    Now differentiate numerator and denominator since you still have $\displaystyle \frac{0}{0}$
    if you try to evaluate the limit.

    Use the product rule for the numerator.
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  10. #10
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    Correct me if I'm wrong, but I believe there's a problem here. The top, $\displaystyle e^{\frac{2}{n}} $ tends towards 1 as $\displaystyle \to \infty$, whereas the bottom $\displaystyle n(1 - e^{\frac{2}{n}}) $ is the indeterminate form $\displaystyle \infty * 0 $ as $\displaystyle n \to \infty $?

    Let's rewrite the limit as $\displaystyle \frac{\mathop{\lim}\limits_{n \to \infty} e^{\frac{2}{n}}}{ \mathop{\lim}\limits_{n \to \infty} n(1 - e^{\frac{2}{n}})} $

    We need to evaluate the bottom first. Change it to

    $\displaystyle \mathop{\lim}\limits_{n \to \infty} \frac{1 - e^{\frac{2}{n}}}{\frac{1}{n}} $

    Applying L'Hopital's rule,

    $\displaystyle \mathop{\lim}{\limits_{n \to \infty} \frac{ \frac{ 2e^{\frac{2}{n}}}{n^2} }{- \frac{1}{n^2}} $.

    Which simplifies to $\displaystyle \mathop{\lim}{\limits_{n \to \infty} 2e^{\frac{2}{n}} = -2$.

    and so our solution is $\displaystyle -2(1 - e^2) \frac{1}{2} = (e^2 - 1)$?
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  11. #11
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    I used the product rule on the numerator and got:

    $\displaystyle x(e^{2x})' + (e^{2x})x'$

    How do I differentiate the denominator?
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  12. #12
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    Math Major, the answer to this problem can be found at Solutions to the Limit Definition of a Definite Integral . The problem was that I didn't understand anything past the L'Hopital's Rule point.
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  13. #13
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    Quote Originally Posted by RogueDemon View Post
    I used the product rule on the numerator and got:

    $\displaystyle x(e^{2x})' + (e^{2x})x'$

    How do I differentiate the denominator?
    $\displaystyle \frac{d}{dx}\left(1-e^{2x}\right)=0-2e^{2x}$

    Now you have $\displaystyle e^{2x}$ as a factor of both numerator and denominator of the result,

    so you end up with $\displaystyle -\frac{1}{2}$

    finally multiply that by the original expression's LH factor.
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  14. #14
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    $\displaystyle \frac{x(e^{2x})' + (e^{2x})x'}{-2(e^{2x})}$

    Will this give $\displaystyle \frac{x + x}{-2}$ ?
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  15. #15
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    Quote Originally Posted by RogueDemon View Post
    $\displaystyle \frac{x(e^{2x})' + (e^{2x})x'}{-2(e^{2x})}$

    Will this give $\displaystyle \frac{x + x}{-2}$ ?
    No!

    u=2x

    $\displaystyle \displaystyle\huge\frac{d}{dx}e^{2x}=\frac{d}{dx}e ^u=\frac{du}{dx}\frac{d}{du}e^u=2e^u=2e^{2x}$

    then

    $\displaystyle \displaystyle\huge\frac{\frac{d}{dx}\left(xe^{2x}\ right)}{\frac{d}{dx}\left(1-e^{2x}\right)}$

    $\displaystyle =\displaystyle\huge\frac{x2e^{2x}+e^{2x}(1)}{-2e^{2x}}$

    $\displaystyle =\displaystyle\huge\frac{e^{2x}\left(2x+1\right)}{ e^2x(-2)}=\frac{2x+1}{-2}$

    Now set x to zero.
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