What is lim_(x->0) (sqrt(x+1)-1)/(root(3)(x+1)-1) ?

1 Answer
Jan 14, 2018

lim_(x->0) (sqrt(x+1)-1)/(root(3)(x+1)-1) = 3/2

Explanation:

Let:

t=root(6)(x+1)

Then:

lim_(x->0) (sqrt(x+1)-1)/(root(3)(x+1)-1) = lim_(t->1) (t^3-1)/(t^2-1)

color(white)(lim_(x->0) (sqrt(x+1)-1)/(root(3)(x+1)-1)) = lim_(t->1) (color(red)(cancel(color(black)((t-1))))(t^2+t+1))/(color(red)(cancel(color(black)((t-1))))(t+1))

color(white)(lim_(x->0) (sqrt(x+1)-1)/(root(3)(x+1)-1)) = lim_(t->1) (t^2+t+1)/(t+1)

color(white)(lim_(x->0) (sqrt(x+1)-1)/(root(3)(x+1)-1)) = 3/2

Alternative method

Alternatively, we can use the binomial theorem to find:

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

color(white)(lim_(x->0) (sqrt(x+1)-1)/(root(3)(x+1)-1)) = lim_(x->0) ((1+1/2x+O(x^2))-1)/((1+1/3x+O(x^2))-1)

color(white)(lim_(x->0) (sqrt(x+1)-1)/(root(3)(x+1)-1)) = lim_(x->0) (1/2x+O(x^2))/(1/3x+O(x^2))

color(white)(lim_(x->0) (sqrt(x+1)-1)/(root(3)(x+1)-1)) = lim_(x->0) (1/2+O(x))/(1/3+O(x))

color(white)(lim_(x->0) (sqrt(x+1)-1)/(root(3)(x+1)-1)) = 3/2