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Thread: L'Hopital's Rule

  1. #1
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    L'Hopital's Rule

    I need to apply L'hopital's rule as x approaches infinity for the following:

    (e^2x)/((x+5)^3)

    which gives me:

    (2e^2x)/((x+5)^3) - (3e^2x)/((x+5)^4)

    and a second time gives me:

    (4e^2x)/((x+5)^3) - (12e^2x)/((X+5)^4) + (12e^2x)/((X+5)^5)


    I guess my question is, how do I know when to stop (i.e. when have I found the limit as x approaches infinity with L'Hopital's rule?)
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  2. #2
    Super Member wingless's Avatar
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    Apply L'hŰpital's Rule until you get something except the indeterminate form. You should also stop when you get $\displaystyle \mp \infty$ or $\displaystyle 0$ too.
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  3. #3
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    Hello, pennydooodle!

    You don't seem to have a grip on L'Hopital's Rule . . .


    Apply L'Hopital's Rule: .$\displaystyle \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3}$
    You seem to be using some kind of strange Quotient Rule here . . .


    Differentiate the numerator: .$\displaystyle 2e^{2x}$

    Differentiate the denominator: .$\displaystyle 3(x+5)^2$

    . . Then: .$\displaystyle \;\lim_{x\to\infty}\frac{2e^{2x}}{3(x+5)^2} \quad\Rightarrow\quad \frac{\infty}{\infty}$


    Apply L'Hopital again.

    $\displaystyle \begin{array}{cc}\text{Numerator:} & 4e^{2x} \\
    \text{Denominator:} & 6(x+5) \end{array} $

    . . Then: .$\displaystyle \lim_{x\to\infty}\frac{4e^{2x}}{6(x+5)} \quad\Rightarrow\quad \frac{\infty}{\infty}$


    Apply L'Hopital again:

    $\displaystyle \begin{array}{cc}\text{Numerator:} & 8e^{2x} \\
    \text{Denominator: } & 6 \end{array}$

    . . Then: .$\displaystyle \lim_{x\to\infty}\frac{8e^{2x}}{6} \quad\Rightarrow\quad \infty $

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  4. #4
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    Thank you Soroban!

    Now I see what I was doing wrong. I was attempting to differentiate the entire equation rather than each part on its own.

    Thank you again!
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