How do you determine the limit of #(sqrt(x^2+10x+1)-x)# as x approaches infinity?

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Answer 1 · 2016-06-28T20:08:33

5

#sqrt(x^2+10x+1)-x#

#= x { (sqrt(1+10/x+1/x^2)-1}#

because #lim_{x \to oo} 10/x+1/x^2 = 0#

.... we can use a Binimial Expansion on #sqrt(1+10/x+1/x^2)# as follows

#sqrt(1+[10/x+1/x^2])#
#= 1 + 1/2 [10/x+1/x^2] + (1/2 (- 1/2))/(2!) [10/x+1/x^2]^2#

#approx 1 + 5/x + mathcal{O} (1/x^2)#

so
#lim_{x to oo} x { (sqrt(1+10/x+1/x^2)-1}#

#= lim_{x to oo} x { 1+5/x-1}#

#=5#