Here is another way it can be done, without using l'Hopital's rule

(I noticed you had another enotes question that required avoiding l'Hopital's rule)

You can factor:

`y^2-1` as `(y-1)(y+1)`

and

`y^3-1` as `(y-1)(y^2+y+1)`

In a similar way, you can write:

`x-1` as

`(sqrt(x)-1)(sqrt(x)+1)` or as

`(root(3)(x)-1)(root(3)(x^2)+root(3)(x)+1)`

So, you can rewrite the limit as:

`(root(3)(x)+sqrt(x) - 2)/(x-1)=`

`(root(3)(x) - 1)/(x-1) + (sqrt(x)-1)/(x-1)=`

`(root(3)(x)-1)/((root(3)(x)-1)(root(3)(x^2)+root(3)(x)+1)) + (sqrt(x)-1)/((sqrt(x)-1)(sqrt(x)+1))=`

`1/(root(3)(x^2)+root(3)(x)+1) + 1/(sqrt(x)+1)`

From here, you can take the limit as `x->1` by just plugging in x=1:

`1/(1+1+1)+1/(1+1) = 5/6`

You need to evaluate the limit, hence, you need to replace 1 for x in equation under limit, such that:

`lim_(x->1) (x^(1/3) + x^(1/2) - 2)/(x - 1) = (1 + 1 - 2)/(1 - 1) = 0/0`

The indetermination case `0/0 ` requests for you to use l'Hospital's theorem, such that:

`lim_(x->1) (x^(1/3) + x^(1/2) - 2)/(x - 1) = lim_(x->1) ((x^(1/3) + x^(1/2) - 2)')/((x - 1)') `

`lim_(x->1) ((x^(1/3) + x^(1/2) - 2)')/((x - 1)') = lim_(x->1) (1/3)*x^(1/3 - 1) + (1/2)x^(1/2 - 1)`

`lim_(x->1) (1/3)*x^(1/3 - 1) + (1/2)x^(1/2 - 1) = lim_(x->1) (1/3)*x^(-2/3) + (1/2)x^(-1/2)`

Using negative power property yields:

`lim_(x->1) (1/3)*x^(-2/3) + (1/2)x^(-1/2) = lim_(x->1) 1/(3root(3)(x^2)) + 1/(2sqrtx) = 1/3 + 1/2 `

`lim_(x->1) 1/(3root(3)(x^2)) + 1/(2sqrtx) = (2+3)/6`

`lim_(x->1) 1/(3root(3)(x^2)) + 1/(2sqrtx) = = 5/6`

**Hence, evaluating the given limit, using l'Hospital's theorem, yields **`lim_(x->1) (x^(1/3) + x^(1/2) - 2)/(x - 1) = 5/6.`

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.