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Thread: Limit of e

  1. #1
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    Limit of e

    Let f be the function given by f(x)=2xe^(2x)

    a) Find Lim as x reaches infinity and negative infinity (i.e. Lim )
    x->infinity/-infinity

    b) What is the range of f?

    Thanks a lot for the help, e always confuses me and i have no idea how i should do this problem.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by aussiekid90 View Post
    Let f be the function given by f(x)=2xe^(2x)

    a) Find Lim as x reaches infinity and negative infinity (i.e. Lim )
    x->infinity/-infinity
    As both $\displaystyle x$ and $\displaystyle e^{2x}$ go to infinity as $\displaystyle x$ goes to $\displaystyle + \infty$ clearly the

    $\displaystyle \lim_{x \to \infty} 2x e^{2x}=\infty$.

    Now as $\displaystyle x \to -\infty,\ x \to -\infty$ and $\displaystyle e^{2x} \to 0$, so we gave an indeterminent form, so we apply L'Hopitals rule

    Put:

    $\displaystyle
    f(x)=\frac{2x}{e^{-2x}}
    $

    then:

    $\displaystyle
    \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{2(-2)}{e^{-2x}}=0
    $

    b) What is the range of f?

    Thanks a lot for the help, e always confuses me and i have no idea how i should do this problem.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by aussiekid90 View Post
    b) What is the range of f?

    Thanks a lot for the help, e always confuses me and i have no idea how i should do this problem.
    We know that $\displaystyle f(x) $ goes to $\displaystyle +\infty$ as $\displaystyle x$ becomes large , and goes to $\displaystyle 0$
    as $\displaystyle x$ goes to $\displaystyle -\infty$. Also differentiating and setting the derivative to zero
    shows that $\displaystyle f(x)$ has a minimum of $\displaystyle -e^{-1}$ at $\displaystyle x=-1/2$.

    So the range of f is $\displaystyle [-e^{-1}, +\infty)$.

    RonL
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