# Limit problem. as x approaches negative infninty. f(x)=x^(2)e^(6x)

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Find `lim_(x-> -oo) x^2e^(6x)` :

For `x<0` this is equivalent to `lim_(x-> -oo) (x^2)/(e^(6x))` or `(lim_(x-> -oo)x^2)/(lim_(x-> -oo)e^(6x))=(-oo)/(oo)` .

This is an indeterminate form. Since both `x^2` and `e^(6x)` are differentiable, by L'Hopital's Rule we have:

`lim_(x->-oo)(x^2)/(e^(6x))=lim_(x->-oo)(2x)/(6e^(6x))` . This limit also yields the indeterminate form `(-oo)/(oo)` ; both numerator and denominator are differentiable so we can apply L'Hopital's Rule again to get:

`lim_(x->-oo)(2x)/(6e^(6x))=lim_(x->-oo)(2)/(36e^(6x))=0` .

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**Therefore**, `lim_(x->-oo)(x^2)/(e^(6x))=0`

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* Note that exponentials grow far faster than any polynomials, so the denominator grows far faster than the numerator supporting the answer. Also consider the graph:

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