Given: #lim_(x to oo) (3x^2+4x)^3/ (4x^3-3)^2#
Evaluation at #oo#, yields the indeterminate form #oo/oo#, therefore, we use L'Hôpital's rule by differentiating the number and the denominator:
#lim_(x to oo) ((d(3x^2+4x)^3)/dx)/ ((d(4x^3-3)^2)/dx)#
#lim_(x to oo) (6 x^2 (3 x + 2) (3 x + 4)^2)/ ( 24 x^2 (4 x^3 - 3))#
It still yields #oo/oo#, therefore, we, again, apply L'Hôpital's rule:
#lim_(x to oo) ((d(6 x^2 (3 x + 2) (3 x + 4)^2))/dx)/ ((d( 24 x^2 (4 x^3 - 3)))/dx)#
#lim_(x to oo) (6 x (135 x^3 + 360 x^2 + 288 x + 64))/(48 x (10 x^3 - 3))#
We repeat this process, until we do not have an indeterminate form:
#lim_(x to oo) ((d(6 x (135 x^3 + 360 x^2 + 288 x + 64)))/dx)/((d(48 x (10 x^3 - 3)))/dx)#
#lim_(x to oo)(24 (135 x^3 + 270 x^2 + 144 x + 16))/(48 (40 x^3 - 3)#
#lim_(x to oo)((d(24 (135 x^3 + 270 x^2 + 144 x + 16)))/dx)/((d(48 (40 x^3 - 3))/dx)#
#lim_(x to oo)(216 (45 x^2 + 60 x + 16))/(5760 x^2)#
#lim_(x to oo)((d(216 (45 x^2 + 60 x + 16)))/dx)/((d(5760 x^2))/dx)#
#lim_(x to oo)(6480 (3 x + 2))/(11520 x)#
#lim_(x to oo)((d(6480 (3 x + 2)))/dx)/((d(11520 x))/dx)#
#lim_(x to oo)(19440)/(11520) = 27/16#
L'Hôpital's rule states that, as goes this limit, so goes the original limit:
#lim_(x to oo) (3x^2+4x)^3/ (4x^3-3)^2 = 27/16#
