What is the limit of #(2x^2+3)^(1/2) - (2x^2-5)^(1/2)# as x approaches infinity?

1 Answer
Apr 11, 2015

The limit is #0#

#((sqrt(2x^2+3) - sqrt(2x^2-5)))/1 ((sqrt(2x^2+3) + sqrt(2x^2-5)))/((sqrt(2x^2+3) + sqrt(2x^2-5))) = ((2x^2+3) - (2x^2-5))/(sqrt(2x^2+3) + sqrt(2x^2-5))#

# = 8/(sqrt(2x^2+3) + sqrt(2x^2-5)) = 8/(x(sqrt(2+3/x^2) + sqrt(2-5/x^2))#

As #x rarr oo#, the denominator also #rarr oo#, so lhe fraction #rarr 0#.

In anticipation:
Any proposed other method needs to also give the correct answers:

#lim_(xrarroo)(sqrt(x^2+3x) - sqrt(x^2+9x)) = -3#

#lim_(xrarr-oo)(sqrt(x^2+3x) - sqrt(x^2+9x)) = 3#