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

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Answer 1 · 2017-02-27T10:36:39

The limit:

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

does not exist since the right and left limit are different.

Around #x=0# we have that: #x+2 > 0#, so #abs(x+2) = x+2#, and then:

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

Evaluate separately:

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

As when #x->0^+# we have #x > 0# so #abs(x) = x#, while:

#lim_(x->0^-) x/abs(x) = -1#

We can conclude that:

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

does not exist since the right and left limit are different.