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

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Answer 1 · 2016-12-29T15:07:54

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

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#