Find Limit of sqrt (8x^2 + 4x - 8) - sqrt (2x^2 - 3x+1) - sqrt (2x^2 + 4x-3) as x->infinity?

how to rationalize

1 Answer
Cesareo R.
Mar 8, 2018

#1/(2 sqrt2)#

Explanation:

#sqrt (8x^2 + 4x - 8) - sqrt (2x^2 - 3x+1) - sqrt (2x^2 + 4x-3) = #

#1/2sqrt (8x^2 + 4x - 8)- sqrt (2x^2 - 3x+1) + 1/2sqrt (8x^2 + 4x - 8)-sqrt (2x^2 + 4x-3)#

Now

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

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

and

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

#(1-3x)/(1/2sqrt (8x^2 + 4x - 8)+sqrt (2x^2 + 4x-3))#

Now

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

#lim_(x->oo)(4-3/x)/(1/2sqrt(8+4/x-8/x^2)+sqrt(2-3/x+1/x^2)) = 4/(2 sqrt2)#

analogously we can proceed obtaining the final result

#lim_(x->oo)sqrt (8x^2 + 4x - 8) - sqrt (2x^2 - 3x+1) - sqrt (2x^2 + 4x-3) = 1/(2 sqrt2)#

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