What is the limit of #(sqrt(x^2+4x+1)-x) # as x goes to infinity?

1 Answer
Jim H
Oct 10, 2015

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

Explanation:

#lim_(xrarroo)(sqrt(x^2+4x+1)-x) # has form #oo-oo# which is indeterminate.

#(sqrt(x^2+4x+1)-x) = ((sqrt(x^2+4x+1)-x))/1 ((sqrt(x^2+4x+1)+x))/((sqrt(x^2+4x+1)+x))#

# = (x^2+4x+1-x^2)/(sqrt(x^2+4x+1)+x)#

# = (4x+1)/(sqrt(x^2(1+4/x+1/x^2))+x)# for #x != 0#

# = (x(4+1/x))/(x(sqrt(1+4/x+1/x^2)+1))# for #x > 0#

# = (4+1/x)/(sqrt(1+4/x+1/x^2)+1)# for #x > 0#

Taking the limit as #xrarroo#, we get #4/(sqrt1+1) = 4/2=2#

Related questions
See all questions in Limits at Infinity and Horizontal Asymptotes
Impact of this question
37700 views around the world
You can reuse this answer
Creative Commons License