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