How do you find the limit of #(2x^3+3x^2cosx)/(x+2)^3# as #x->oo#?

1 Answer
Oct 23, 2016

The limit has indeterminate initial form #oo/oo#. We'll try to make the denominator go to someting oher than #oo#.

Explanation:

We want the limit as #xrarroo#, we aren't worried about what happens at #0#.

For #x != 0#,

#(2x^3+3x^2cosx)/(x+2)^3 = (x^3(2+(3cosx)/x))/(x(1+2/x))^3#

# = (x^3(2+(3cosx)/x))/(x^3(1+2/x)^3)#

# = (2+(3cosx)/x)/(1+2/x)^3#

As #xrarroo#, the denominator goes to #1# and the numerator goes to #2#

So the limit of the ratio as #xrarroo# is #2#