What is the limit of #sqrt(x^2+8x-5)-sqrt(x^2-6x+2)# as x approaches infinity?

1 Answer
Cesareo R.
Aug 24, 2016

#7#

Explanation:

#(sqrt(x^2+8x-5)-sqrt(x^2-6x+2))(sqrt(x^2+8x-5)+sqrt(x^2-6x+2))/(sqrt(x^2+8x-5)+sqrt(x^2-6x+2)) =#
#=(x^2+8x-5-(x^2-6x+2))/(sqrt(x^2+8x-5)+sqrt(x^2-6x+2))=#
#=(14x-7)/(sqrt(x^2+8x-5)+sqrt(x^2-6x+2)) =#
#=(14x-7)/(x(sqrt(1+8/x-5/x^2)+sqrt(1-6/x+2/x^2))) =#
#(14-7/x)/(sqrt(1+8/x-5/x^2)+sqrt(1-6/x+2/x^2))#

Finally

#lim_{x->oo}(sqrt(x^2+8x-5)-sqrt(x^2-6x+2))=14/2=7#

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