Question

How do I find the limit of this ?

`lim_(x to 1) (root(3)(x^2)-2(root(3)(x))+1)/(x-1)^2` can you please explain this in detail ?

Answer 1

Because the expression evaluated at `x = 1` yields the indeterminate form `0/0`, one should use L'Hôpital's rule .

Explanation:

Given:
`lim_(x to 1) (root(3)(x^2)-2(root(3)(x))+1)/(x-1)^2`

Because we are about to differentiate the numerator and denominator, I shall write the radicals as powers:

`lim_(x to 1) (x^(2/3)-2x^(1/3)+1)/(x-1)^2`

To apply L'Hôpital's rule, we differentiate the numerator and the denominator:

`lim_(x to 1) ((d(x^(2/3)-2x^(1/3)+1))/dx)/((d(x-1)^2)/dx)`

Compute the derivatives:

`lim_(x to 1) (2/3x^(-1/3)-2/3x^(-2/3))/(2(x-1))`

A common factor of 2 cancels:

`lim_(x to 1) (1/3x^(-1/3)-1/3x^(-2/3))/(x-1)`

This, also, yields the indeterminate form of `0/0`, therefore, we apply L'Hôpital's rule, again:

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

Compute the derivatives:

`lim_(x to 1) (-1/9x^(-4/3)+2/9x^(-5/3))/1`

We can evaluate this at `x = 1`:

`-1/9(1)^(-4/3)+2/9(1)^(-5/3) = 1/9`

L'Hôpital's rule states that, as goes the above limit, so goes the original limit:

`lim_(x to 1) (root(3)(x^2)-2(root(3)(x))+1)/(x-1)^2 = 1/9`