How do you find #lim (sqrt(x+2)-sqrtx)/(sqrt(x+3)-sqrtx)# as #x->oo#?

1 Answer
Cesareo R.
Feb 11, 2017

#2/3#

Explanation:

#(sqrt(x+2)-sqrt(x))/(sqrt(x+3)-sqrt(x)) = (sqrt(x+2)-sqrt(x))/(sqrt(x+3)-sqrt(x)) ((sqrt(x+3)+sqrt(x))/(sqrt(x+3)+sqrt(x)))#
#=((sqrt(x+2)-sqrt(x))(sqrt(x+3)+sqrt(x)) )/3#
Calling now #f(x)=((sqrt(x+2)-sqrt(x))(sqrt(x+3)+sqrt(x)) )/3# we have

#f(x)((sqrt(x+2)+sqrt(x))/(sqrt(x+2)+sqrt(x))) = #
#2/3((sqrt(x+3)+sqrt(x)) /(sqrt(x+2)+sqrt(x)))=(2/3)(sqrt(1+3/x)+1)/(sqrt(1+2/x)+1)#

Finally

#lim_(x->oo)(sqrt(x+2)-sqrt(x))/(sqrt(x+3)-sqrt(x))=lim_(x->oo)(2/3)(sqrt(1+3/x)+1)/(sqrt(1+2/x)+1) = 2/3#

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