Question

What is `lim_(x->1) (x^(1/3)-1)/(x^(1/4)-1)` ?

Answer 1

`4/3`

Explanation:

`(x^(1/3)-1)/(x^(1/4)-1)` Plugging in 1 gives the indeterminate form `0/0`

Using L'Hospital's rule:

`d/dx(x^(1/3)-1)=1/(3x^(2/3))`

`d/dx(x^(1/3)-1)=1/(4x^(3/4)`

`(1/(3x^(2/3)))/(1/(4x^(3/4)))=(4x^(3/4))/(3x^(2/3))`

Plugging in 1:

`(4x^(3/4))/(3x^(2/3))= (4(1)^(3/4))/(3(1)^(2/3))=4/3`

So:

`lim_(x->1)(x^(1/3)-1)/(x^(1/4)-1)=4/3`

Answer 2

`lim_(x->1) (x^(1/3)-1)/(x^(1/4)-1) = 4/3`

Explanation:

Let `x=t^12`

Then:

`lim_(x->1) (x^(1/3)-1)/(x^(1/4)-1) = lim_(t->1) (t^4-1)/(t^3-1)`

`color(white)(lim_(x->1) (x^(1/3)-1)/(x^(1/4)-1)) = lim_(t->1) (color(red)(cancel(color(black)((t-1))))(t^3+t^2+t+1))/(color(red)(cancel(color(black)((t-1))))(t^2+t+1))`

`color(white)(lim_(x->1) (x^(1/3)-1)/(x^(1/4)-1)) = lim_(t->1) (t^3+t^2+t+1)/(t^2+t+1)`

`color(white)(lim_(x->1) (x^(1/3)-1)/(x^(1/4)-1)) = 4/3`

graph{(y-(x^(1/3)-1)/(x^(1/4)-1))((x-1)^2+(y-4/3)^2-0.003) = 0 [-1.32, 3.68, -0.23, 2.27]}

Answer 3

`4/3.`

Explanation:

Here is another way to solve the Problem, without using

L'Hospital's Rule.

We will use the following Standard Form of Limit :

`L : lim_(x to a) (x^n-a^n)/(x-a)=na^(n-1).`

Now, the Reqd. Lim. `=lim_(x to 1)(x^(1/3)-1)/(x^(1/4)-1),`

`=lim{(x^(1/3)-1^(1/3))/(x-1)}-:{(x^(1/4)-1^(1/4))/(x-1)},`

`={lim(x^(1/3)-1^(1/3))/(x-1)}-:{lim(x^(1/4)-1^(1/4))/(x-1)},`

`={1/3*1^(1/3-1)}-:{1/4*1^(1/4-1)},`

`=1/3-:1/4.`

`rArr" The Reqd. Lim.="4/3.`

Enjoy Maths.!