# Determine the limit `lim_(x->1)(root(5)(x)-1)/(root(3)(x)-1)`

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The limit `lim_(x->1)(root(5)(x)-1)/(root(3)(x)-1)` has to be determined.

Substituting x = 1 gives the indeterminate form `0/0` . In this case l'Hopital's rule can be used and the numerator and denominator substituted by their derivatives.

`lim_(x->1)(root(5)(x)-1)/(root(3)(x)-1)`

=> `lim_(x->1)((root(5)(x)-1)')/((root(3)(x)-1)')`

=> `lim_(x->1)((1/5)x^(-4/5))/((1/3)*x^(-2/3))`

=> `lim_(x->1)(3/5)(x^(-4/5)/(x^(-2/3)))`

substituting x = 1 gives `3/5`

**The limit `lim_(x->1)(root(5)(x)-1)/(root(3)(x)-1) = 3/5` **

The terms of the limit are not clearly specified, hence, considering `sqrt 5[ x]` = `sqrt(5x)` and `sqrt3[(x)] = sqrt(3x)` , yields:

`lim_(x->1) (sqrt(5x) - 1)/(sqrt(3x) - 1)`

You need to susbtitute 1 for x such that:

`lim_(x->1) (sqrt(5x) - 1)/(sqrt(3x) - 1) = (sqrt 5 - 1)/(sqrt 5 - 1)`

If you mean by sqrt the radical sign, hence, you may consider `sqrt5[(x)] = root(5)(x)` and `sqrt3[(x)] = root(3)(x)` , such that:

`lim_(x->1) (root(5)(x) - 1)/(root(3)(x) - 1) = (root(5)(1) - 1)/(root(3)(1) - 1)`

`lim_(x->1) (root(5)(x) - 1)/(root(3)(x) - 1) = (root(5)(1) - 1)/(root(3)(1) - 1) = (1-1)/(1-1) = 0/0`

Since the limit is indeterminate, you may use l'Hospital's theorem, such that:

`lim_(x->1) (root(5)(x) - 1)/(root(3)(x) - 1) = lim_(x->1) ((root(5)(x) - 1)')/((root(3)(x) - 1)') `

`lim_(x->1) (root(5)(x) - 1)/(root(3)(x) - 1) = lim_(x->1) (1/(5root(5)(x^4)))/(1/(3root(3)(x^2)))`

`lim_(x->1) (root(5)(x) - 1)/(root(3)(x) - 1) = (1/5)/(1/3)`

`lim_(x->1) (root(5)(x) - 1)/(root(3)(x) - 1) = 3/5`

**Hence, evaluating the limit if and , yields `lim_(x->1) (sqrt(5x) - 1)/(sqrt(3x) - 1) = (sqrt 5 - 1)/(sqrt 5 - 1)` and evaluating the limit if `sqrt5[(x)] = root(5)(x)` and `sqrt3[(x)] = root(3)(x)` yieldsl`lim_(x->1) (sqrt(5x) - 1)/(sqrt(3x) - 1) = 3/5.` **