Find the limit?

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

3 Answers
Apr 1, 2018

Answer:

#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#

Apr 1, 2018

Answer:

#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]}

Apr 1, 2018

Answer:

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#