How do you evaluate the limit #(sqrt(x+6)-x)/(x-3)# as x approaches #3#?

1 Answer
Jim G.
Feb 6, 2017

#-5/6#

Explanation:

As the function is indeterminate #0/0# when x = 3

Multiply the numerator/denominator by the conjugate of the numerator.

#sqrt(x+6)-x" conjugate "tosqrt(x+6)+x#

#rArr((sqrt(x+6)-x)(sqrt(x+6)+x))/((x-3)(sqrt(x+6)+x)#

#=(x+6-x^2)/((x-3)(sqrt(x+6)+x))#

#=(-cancel((x-3))(x+2))/(cancel((x-3))(sqrt(x+6)+x))#

exclusion x ≠ 3

#=(-(x+2))/(sqrt(x+6)+x)#

#rArrlim_(xto3)(sqrt(x+6)-x)/(x-3)#

#=lim_(xto3)(-(x+2))/(sqrt(x+6)+x)=-5/6#

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
18691 views around the world
You can reuse this answer
Creative Commons License