Question

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

Answer 1

`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`

Answer 2

`-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.`