How do you evaluate the limit #(sqrt(x+2)-sqrt(2-x))/x# as x approaches #0#?

Question

27852 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-16T13:43:19

#sqrt2/2#.

Reqd. Limit#=lim_(xrarr0) (sqrt(x+2)-sqrt(2-x))/x#

#lim_(xrarr0) (sqrt(x+2)-sqrt(2-x))/x xx (sqrt(x+2)+sqrt(2-x))/(sqrt(x+2)+sqrt(2-x))#

#lim_(xrarr0) (x+2-2+x)/(x(sqrt(x+2)+sqrt(2-x))#

#lim_(xrarr0) (2cancelx)/(cancelx(sqrt(x+2)+sqrt(2-x))#

#=2/(sqrt2+sqrt2)#

#=2/(2sqrt2)#

#=sqrt2/2#.