How do you find the limit of #(2-x)/(x^2-4)# as #x->2#?

Question

1194 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-14T17:35:05

#lim_(xrarr2)(2-x)/(x^2-4) = lim_(xrarr2)(-1)/(x+2) = -1/4#

#2-x = -(x-2)#

and

#x^2-4 = (x+2)(x-2)#.

So for all #x# except #2#, we have

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

The limit doesn't care what happens when #x=2#, just what happens when #x# is close to #2#.

Answer 2 · 2016-11-14T17:42:01

#(-1)/4#

#Lim_( x->2) (2-x)/(x^2-4) = Lim_ (x->2) (2-x)/((x-2)(x+2))#

= #Lim_(x->2) (-1)/(x+2) = -1/4#