lim (x->infinity) of xe^(-x^2)?

csmajor86

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Feb 4, 2008
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I need to evaluate the limit:

lim[sub:pyd78lgy]x->infinity[/sub:pyd78lgy] xe^(-x^2) .

I am trying to approach it using l'Hospital's Rule. I first put the e[sup:pyd78lgy]-x^2[/sup:pyd78lgy] in the denominator to get
x/e[sup:pyd78lgy]x^2[/sup:pyd78lgy] . I can do that right?

This of the "infinity"/"infinity" but after taking the derivative of the top and the bottom I get:

lim[sub:pyd78lgy]x->infinity[/sub:pyd78lgy] 1/2xe[sup:pyd78lgy]x^2[/sup:pyd78lgy] = 0.

Is that correct or am I making a mistake somewhere?
 
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