How do you find the limit #e^x/x^3# as #x->oo#?

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Answer 1 · 2016-12-30T11:47:16

#\lim_(x to oo) (e^x)/(x^3) to oo #

This is in indeterminate #oo/oo# form so you could plod through L'Hôpital's Rule 3 times.

I think it's more revealing to look at the exponential #e^z# function which can be defined as

#e^x =\sum _{k=0}^{\infty } x^{k} / (k!) =1+x+x^2 / 2+x^{3} / 6 + x^4 / 24 + O(x^5) #

So we have

#\lim_(x to oo) (1+x+x^2 / 2+x^{3} / 6 + x^4 / 24 + O(x^5))/(x^3) #

#= \lim_(x to oo) 1/x^3+1/x^2+1 /( 2x) + 1 / 6 + x / 24 + O(x^2)#

First three terms go to zero,

# \lim_(x to oo) 1/x^2, 1/x^2, 1 /( 2x) = 0#

the fourth is constant

# \lim_(x to oo) 1 / 6 = 1/6#

... but look at the fifth and subsequent terms....

#= \lim_(x to oo) x / 24 + O(x^2) to oo#