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

1 Answer
George C.
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#

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