You need to substitute `oo` for x in equation under limit such that:

`lim_(x-gtoo)(14x^2)/(e^(9x)) = oo/oo`

You need to use l'Hospital's theorem such that:

`lim_(x-gtoo)((14x^2)')/((e^(9x))') = lim_(x-gtoo)(28x)/(9e^(9x))`

Substituting `oo ` for x yields an indetermination `oo/oo` , hence you need to use l'Hospital's theorem again such that:

`lim_(x-gtoo)(28x)/(9e^(9x)) = lim_(x-gtoo)((28x)')/((9e^(9x))')`

`lim_(x-gtoo)((28x)')/((9e^(9x))')= lim_(x-gtoo) 28/(81e^(9x))`

`lim_(x-gtoo) 28/(81e^(9x)) = (28/81)*lim_(x-gtoo) 1/(e^(9x))`

`lim_(x-gtoo) 28/(81e^(9x)) = (28/81)*(1/oo)`

`lim_(x-gtoo) 28/(81e^(9x)) = (28/81)*(0)`

`lim_(x-gtoo) 28/(81e^(9x)) = 0`

**Hence, evaluating the limit to the given function yields `lim_(x-gtoo)(14x^2)/(e^(9x)) = 0.` **

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