What is the #lim (sqrt(x^2+3x)-x)#, as x approaches infinity?

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Answer 1 · 2016-10-14T18:04:14

#3/2#

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

#= lim_(x to oo) x sqrt(1+3/x)-x#

by binomial expansion of the radical

#= lim_(x to oo) x (1+1/2*3/x + O(1/x)^2) - x#

#= lim_(x to oo) x+3/2 + O(1/x) - x#

#= lim_(x to oo) 3/2 + O(1/x) = 3/2#

Answer 2 · 2016-10-14T19:51:49

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

# = (3x)/(sqrt(x^2+3x)+x)#

As #xrarroo#, we have #x > 0# so #sqrt(x^2) = x#

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

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

# = (3x)/(x(sqrt(1+3/x)+1))#

# = 3/(sqrt(1+3/x)+1)#

Evaluating limit as #xrarroo#, we get

# 3/2#