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

mlehuzzah | Certified Educator

calendarEducator since 2012

starTop subjects are Math and Literature

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`

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## Related Questions

Luca B. | Certified Educator

calendarEducator since 2011

starTop subjects are Math, Science, and Business

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.`

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