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.

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