# evaluate the limit as x approaches infinity. f(x)=(e^5x)/(x^3)

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``This might look like we need to use L'Hospital's rule three times but

`lim_(x->oo) e^x/x = lim_(x->oo) e^x/1 = lim_(x->oo) e^x = oo`

So since `lim_(x->oo) e^(5x)/x^3 = lim_(x->oo) e^x/x * lim_(x->oo) e^x/x * lim_(x->oo) e^x/x * lim_(x->oo) e^(2x)`

And this has to `= oo` we get

`lim_(x->oo) e^(5x)/x^3 = oo`