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

Question

Charles Arocha · Accepted Answer

#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.

Isaiah Allen · Answer

undefined To find the limit of (x - 3) / (abs(x - 3)) as x approaches 3, we can evaluate the limit from both the left and right sides of 3 separately.

When x approaches 3 from the left side (x < 3), the expression simplifies to (x - 3) / (-(x - 3)), which further simplifies to -1.

When x approaches 3 from the right side (x > 3), the expression simplifies to (x - 3) / (x - 3), which further simplifies to 1.

Since the limit from the left side is -1 and the limit from the right side is 1, the overall limit as x approaches 3 does not exist.