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    i have to prove this Qn:

    Prove; lim_{x-> - \infty}x^{n}e^{x}=0 for all integers x>0.

    I don't know where to start with this, as the limit says as x-> MINUS infinity, but x>0?

    (EDIT)

    any clues please?
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    Prove what?

    Btw, presumably n is the integer not x.
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    Ooops! sorry! that the limit is equal to zero

    (my bad!)
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    (Original post by DrSheldonCooper)
    i have to prove this Qn:

    Prove; lim_{x-> - \infty}x^{n}e^{x}=0 for all integers x>0.

    I don't know where to start with this, as the limit says as x-> MINUS infinity, but x>0?

    (EDIT)

    any clues please?
    Well you should know that \lim_{x \rightarrow- \infty}x^{n} \equiv \lim_{x \rightarrow \infty} \frac{1}{x^n}
    as  x^-^n = \frac{1}{x^n} .
    Now what happens?
    You can do the same thing for  e^n
    Once you know what happens to both of the two it should be obvious why the limit is 0.
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    (Original post by DrSheldonCooper)
    ...
    \displaystyle \begin{aligned} x \mapsto -x \ \Rightarrow \lim_{x \to -\infty} x^k e^x & = \lim_{x \to \infty} (-x)^k e^{-x} \\ & = \lim_{x \to \infty} \frac{(-x)^k}{e^x}

    Consider the series expansion of e^x and see if you can simplify the fraction.
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    It's a bit difficult to give you an answer to this question without knowing what kind of proof you expect: epsilon/delta or is faffing about with "lim" notation enough? How have you defined exp?
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    Oh I just thought of an alternative method. It requires knowledge of L'Hopital's rule:
 
 
 

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