How do you find the limit of #((x^2 + 2x - 3)^0.5) - x# as x approaches #oo#?

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Answer 1 · 2018-04-12T08:40:34

#1#

#lim_(x to oo) sqrt(x^2 + 2x - 3) - x#

Complete the square

#= lim_(x to oo) sqrt((x+1)^2- 4) - x#

let #z = x + 1#

#= lim_(z to oo) sqrt(z^2- 4) - z + 1#

#= lim_(z to oo) z sqrt(1- 4/z^2) - z + 1#

Binomial expansion for the root

#= lim_(z to oo) z (1- 2/z^2 + mathbb O (1/z^4) ) - z + 1#

#= lim_(z to oo) z- 2/z + mathbb O (1/z^3) - z + 1 = 1#