# Finding limits with absolute values

1. Oct 15, 2006

### Alain12345

I need help finding limits. I know it's pretty simple most of the time... I know that for example if the lim x--> 3 of x-3, you just plug 3 for x... what do I do if it's the absolute value of x-3?

I know you guys like to see that I've tried to solve the problem, but there's not much I can show you in this case.

Thanks.

2. Oct 15, 2006

you have to look at the right and left hand limits (i.e. lim as x approaches a from the left, etc...)

Last edited: Oct 15, 2006
3. Nov 9, 2006

### lonelywizard

If you have an expression that is in indeterminate form, you may try to factor it out. If you cannot factor it, then try multiply it with its conjugate pair.

4. Nov 10, 2006

### drpizza

For absolute value problems, you can write it as a piecewise defined function. i.e. instead of y = abs(x), you can write y = { x, x>=0; -x,x<0

Then, as courtrigard said, look at the left hand and right hand limits.

5. Oct 6, 2010

### defansimoh

Finding d limit of the absolute value of x-3, divided by x-3, for x approuches 0.

lim [x-3]/(x-3)
x->0
Principle [x-3]= {(x-3) if x>4}
{-(x-3) if x<4}

Remember that limit olny exist only if the limit from the negetive side is equal to the limit from the positive side as x approuches the given value!

= lim (x-3)/(x-3) = 1 and =lim -(x-3)/(x-3) = -1
x=>0+ x=>0-
we can conclude by saying
1 is not equal to -1
The limit of [x-3]/x-3 as x=>0 does not exist!!!