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Math Help - líHopitalís rule

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    líHopitalís rule

    Find \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3}. Apply líHopitalís rule as many times as necessary, verifying your results after each application
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by FLTR View Post
    Find \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3}. Apply líHopitalís rule as many times as necessary, verifying your results after each application
    \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3}

    First both the numerator and denominator go to infinity in the limit, so we can apply L'Hopital's rule.

    Thus:
    \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3} = \lim_{x\to\infty}\frac{2e^{2x}}{3(x+5)^2}

    Again, both numerator and denominator go to infinity so:
    \lim_{x\to\infty}\frac{2e^{2x}}{3(x+5)^2} = \lim_{x\to\infty}\frac{4e^{2x}}{6(x+5)}

    Again, both numerator and denominator go to infinity so:
    \lim_{x\to\infty}\frac{4e^{2x}}{6(x+5)} = \lim_{x\to\infty}\frac{8e^{2x}}{6}

    Finally, only the numerator goes to infinity and the denominator is a constant, so the net result is:
    \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3} = \lim_{x\to\infty}\frac{8e^{2x}}{6} \to + \infty

    or more precisely, the limit is undefined.

    -Dan
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    Senior Member TriKri's Avatar
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    Just another question, a little of topic maybe, how can one explain so I understand why líHopitalís rule is true when f(c)=g(c)=\infty? When f(c)=g(c)=0 it is easier.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TriKri View Post
    Just another question, a little of topic maybe, how can one explain so I understand why líHopitalís rule is true when f(c)=g(c)=\infty? When f(c)=g(c)=0 it is easier.
    All that my texts will say on the matter is that the proof is "technical." I assume that to be on the same level as "non-trivial" which is usually rather nasty.

    -Dan
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    Quote Originally Posted by TriKri View Post
    Just another question, a little of topic maybe, how can one explain so I understand why líHopitalís rule is true when f(c)=g(c)=\infty? When f(c)=g(c)=0 it is easier.
    Proof.
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