Question

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

Answer 1

`lim_(x->+oo) (e^x-x^2) = +oo`

Explanation:

You can write `f(x)` as:

`e^x-x^2 = x^2(e^x/x^2-1)`

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->+oo) x^2(e^x/x^2-1) = +oo`