# How do you evaluate the limit lim e^x-x^2 as x->oo?

Dec 29, 2016

${\lim}_{x \to + \infty} \left({e}^{x} - {x}^{2}\right) = + \infty$

#### Explanation:

You can write $f \left(x\right)$ as:

${e}^{x} - {x}^{2} = {x}^{2} \left({e}^{x} / {x}^{2} - 1\right)$

now develop ${e}^{x}$ in its MacLaurin series:

e^x= sum_(n=0)^oo x^n/(n!)=1+x+x^2/(2!)+x^3/(3!)+...

then you can see that:

(e^x/x^2-1) = 1/x+1/2+x/(3!)+...=1/x+1/2+sum_(n=3)^oo x^(n-2)/(n!)

As the series is absolutely convergent, we have that:

lim_(x->+oo)(e^x/x^2-1) = lim_(x->+oo) 1/x+1/2+sum_(n=0)^oo [lim_(x->+oo)(x^(n-2)/(n!))] =+oo

and therefore:

${\lim}_{x \to + \infty} {x}^{2} \left({e}^{x} / {x}^{2} - 1\right) = + \infty$