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Answer 1 · 2018-01-19T23:31:24

#3/2#

#sqrt(4x^2+8x)-sqrt(x^2+1)-sqrt(x^2+x) =#

#=sqrt(x^2+2x)-sqrt(x^2+1)+sqrt(x^2+2x)-sqrt(x^2+x) =#

#= ((x^2+2x)-(x^2+1))/(sqrt(x^2+2x)+sqrt(x^2+1))+((x^2+2x)-(x^2+x))/(sqrt(x^2+2x)+sqrt(x^2+x)) =#

#=(2x-1)/(sqrt(x^2+2x)+sqrt(x^2+1))+x/(sqrt(x^2+2x)+sqrt(x^2+x)) =#

#(2-1/x)/(sqrt(1+2/x)+sqrt(1+1/x^2))+1/(sqrt(1+2/x)+sqrt(1+1/x)) #

and then

#lim_(x->oo)sqrt(4x^2+8x)-sqrt(x^2+1)-sqrt(x^2+x) =2/2+1/2 = 3/2#

Answer 2 · 2018-01-20T11:53:02

Answer to the question: "What is the limit as x approaches infinity of #sqrt(4x^2+8x)-sqrt(x^2+1)-sqrt(x^2+x)?#

#3/2#

#sqrt(4x^2+8x)-sqrt(x^2+1)-sqrt(x^2+x) =#

#x(2sqrt(1+2/x)-sqrt(1+1/x^2)-sqrt(1+1/x))#

now calling #y = 1/x# we have

#lim_(x->oo)sqrt(4x^2+8x)-sqrt(x^2+1)-sqrt(x^2+x)equiv lim_(y->0)1/y(2 sqrt(1+2y)-sqrt(1+y^2)-sqrt(1+y))#

now developing #2 sqrt(1+2y)-sqrt(1+y^2)-sqrt(1+y)# in series aroung #y=0# we get

#2 sqrt(1+2y)-sqrt(1+y^2)-sqrt(1+y) approx (3 y)/2 - (11 y^2)/8 + (15 y^3)/16 - (139 y^4)/128+O(y^5)#

and then

#lim_(y->0)1/y(2 sqrt(1+2y)-sqrt(1+y^2)-sqrt(1+y))=#

#lim_(y->0) (3 )/2 - (11 y)/8 + (15 y^2)/16 - (139 y^3)/128+O(y^3) = 3/2#