Find the value of #lim ((cosx)^(1/2)-(cosx)^(1/3))/sin^2x# as #x# approaches #0#?

2 Answers
Dec 6, 2017

#lim_(x→0)(cos^(1/2)x-cos^(1/3)x)/sin^2x=-1/12#

Explanation:

The first thing to do when given a limit is to plug the limiting value into the expression and see what happens.

#(cos^(1/2)(0)-cos^(1/3)(0))/sin^2(0)=(1-1)/0=0/0#

Since the expression is indeterminate, we may use l'Hôpital's rule.

#lim_(x→0)(cos^(1/2)x-cos^(1/3)x)/sin^2x=lim_(x→0)(d/dx(cos^(1/2)x-cos^(1/3)x))/(d/dx(sin^2x))=lim_(x→0)(-sinx/(2cos^(1/2)x)+sinx/(3cos^(1/3)x))/(2sinxcosx)=lim_(x→0)(2cos^(1/2)x-3cos^(1/3)x)/(12cos^(11/6)x)=(2cos^(1/2)(0)-3cos^(1/3)(0))/(12cos^(11/6)(0))=(2-3)/12=-1/12#

Dec 6, 2017

# -1/12#.

Explanation:

Here is another Method to find the required Limit L :

We use the substituion #cosx=y^6.#

# :." As "x to 0, y to 1.#

#:. L=lim_(x to 0) {(cosx)^(1/2)-(cosx)^(1/3)}/sin^2x.#

Using #sin^2x=1-cos^2x,# we have,

#=lim_(y to 1){(y^6)^(1/2)-(y^6)^(1/3)}/{1-(y^6)^2},#

#=lim(y^3-y^2)/(1-y^12),#

#=lim{y^2(y-1)}/{(1+y^6)(1-y^6)},#

#=lim{y^2(y-1)}/{(1+y^6)(1+y^3)(1-y^3)},#

#=lim{y^2cancel((y-1))}/{(1+y^6)(1+y^3)cancel((1-y))(1+y+y^2)},#

#=lim-y^2/{(1+y^6)(1+y^3)(1+y+y^2)},#

#=-1/{(1+1)(1+1)(1+1+1)}.#

# rArr L=-1/12.#