How do you evaluate the limit #(1/(x+2)-1/2)/x# as x approaches #0#?

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Answer 1 · 2016-11-10T14:15:01

The limit is #-1/4#.

#=lim_(x->0) ((2 - (x + 2))/(2(x + 2)))/x#

#=lim_(x->0) ((2 - x - 2)/(2(x+ 2)))/x#

#=lim_(x->0) ((-x)/(2(x+ 2)))/x#

#=lim_(x->0) (-x)/(2(x + 2)) xx 1/x#

#=lim_(x->0) -1/(2(x + 2))#

#= -1/(2(0 + 2))#

#=-1/4#

Hopefully this helps!