How do you find the limit of # (x - 3) / (abs(x - 3))# as x approaches 3?

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Answer 1 · 2016-07-21T18:23:55

#lim_(xrarr3)f(x)# does not exist.

Recall that, #|y|=y, y>=0, and, |y|=-y, if y<0#.

Let us denote, by #f(x)=(x-3)/|x-3|, x!=3#.

As #xrarr3+, x>3 rArr (x-3)>0rArr |x-3|=x-3#

#rArr f(x)=(x-3)/|x-3|=(x-3)/(x-3)=1, as, x!=3#.

#:. lim_(xrarr3+)f(x)=lim_(xrarr3+) 1=1...........(1)#

As #xrarr3-, x<3 rArr (x-3)<0rArr|x-3|=-(x-3)#

#rArr f(x)=(x-3)/|x-3|=(x-3)/-(x-3)=-1, as x!=3#.

#:. lim_(xrarr3-)f(x)=lim_(xrarr3-)-1=-1..............(2)#

From #(1) and (2)#, we see that,

#lim_(xrarr3+)f(x)=1!=-1=lim_(xrarr3-)f(x)#

Evidently, #lim_(xrarr3)f(x)# does not exist.