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Math Help - Can someone verify this L'Hospital problem

  1. #1
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    Can someone verify this L'Hospital problem

    \lim \rightarrow \infty \frac{e^{-5x}}{e^{3x}}

    \lim \rightarrow \infty \frac{-5e^{-5x}}{3e^{3x}}

    \lim \rightarrow \infty \frac{-5}{3e^{3x}e^{5x}}=-5
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  2. #2
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    Quote Originally Posted by redsox25 View Post
    \lim \rightarrow \infty \frac{e^{-5x}}{e^{3x}}

    \lim \rightarrow \infty \frac{-5e^{-5x}}{3e^{3x}}

    \lim \rightarrow \infty \frac{-5}{3e^{3x}e^{5x}}=-5

    Any positive number raised to any power is always positive ==> the limit cannot be a negative number.

    When x --> oo, e^(-3x) --> 0 and e^5x --> oo, so you can't uise L'Hospital as you did.

    The limit is e^(-3x)*1/(e^(5x)) --> 0*0 = 0

    Tonio
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  3. #3
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    Quote Originally Posted by redsox25 View Post
    \lim \rightarrow \infty \frac{e^{-5x}}{e^{3x}}

    \lim \rightarrow \infty \frac{-5e^{-5x}}{3e^{3x}}

    \lim \rightarrow \infty \frac{-5}{3e^{3x}e^{5x}}=-5
    \lim_{x \to \infty} \frac{e^{-5x}}{e^{3x}}

    you can't use L'Hopital in this case.
    numerator goes to 0 , denominator goes to \infty

    try re-writing the expression ...

    \lim_{x \to \infty} \frac{1}{e^{8x}} = 0
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