Question

Find the limit?

`lim_(x rarr -1)(1+root(3)x)/(1+root(5)x`

Answer 1

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

Explanation:

Given: `lim_(x rarr -1)(1+root(3)x)/(1+root(5)x)`

Because the expression evaluated at the limit, yields the indeterminant form, `0/0`, one should use L'Hôpital's rule.

To apply L'Hôpital's rule, one differentiates the numberator and the denominator with respect to the dependent variable:

`lim_(x to -1)(1+root(3)x)/(1+root(5)x) = lim_(x to -1)((d(1+root(3)x))/dx)/((d(1+root(5)x))/dx)`

Perform the differentiations:

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

Simplify

`lim_(x to -1)(1+root(3)x)/(1+root(5)x) = lim_(x to -1) 5/3x^(-2/15)`

The limit on the right can be evaluated at `x = -1`:

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

Answer 2

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

Explanation:

The limit:

`lim_(x->-1) (1+x^(1/3))/(1+x^(1/5))`

is in the indeterminate form `0/0` so we can use l'Hospital's rule:

`lim_(x->-1) (1+x^(1/3))/(1+x^(1/5)) = lim_(x->-1) (d/dx (1+x^(1/3)))/(d/dx (1+x^(1/5)))`

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

graph{(1+root(3)x)/(1+root(5)x) [-2, 0, -1, 5]}

Answer 3

See below

Explanation:

As the root is odd we will handle only the real root.

Making `y = root(15)(x)` we have

`(1+root(3)(x))/(1+root(5)(x)) = (1+y^3)/(1+y^5) =`

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

and finally

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