# Limit Problem Difference of Squares

1. Sep 30, 2007

### Destrio

how would you do:
lim x->3+ of abs(x+3) / x^2 - 9
because if you take the difference of squares of the bottom
and divide x+3 / x+3 for the positive abs case
you are left with 1/(x-3) with x approaching 3
making the denominater 0

I have since realized that the question was asking for -3+, and was able to solve it
but if it was asking for +3+ how could I go about solving it?

Thanks

2. Sep 30, 2007

### Dick

The limit as x->3+ doesn't exist. The numerator approaches 6 and the denominator approaches 0. How could it?

3. Sep 30, 2007

### HallsofIvy

Are you serious? If x= 3, what is |x+3|/(x^2- 9)?

4. Sep 30, 2007

### Destrio

If x= 3, what is |x+3|/(x^2- 9)?
6/0

i know the limit as x approaches 3 doesnt exist
but so far in my course we have always been able to arrange one-sided limits so that they exist

so if it is simplified like that and the denominator remains 0, is it safe to say that is doesnt exist since it is not defined?

5. Sep 30, 2007

### HallsofIvy

No, the limit does not exist!

6. Sep 30, 2007

### Dick

The value of the function at -3+ isn't defined either, but the limit exists. Think of what happens to the values of the function as the limit is approached rather than rather than what happens AT the point.

7. Sep 30, 2007

### dynamicsolo

lim x->-3+ |x+3| / (x^2 - 9)

The one-sided limits do exist: they just don't match up.

For x-> -3+ , |x+3| = (x+3) , so the expression will reduce to
1/(x-3) . Then the limit is 1/-6 = -1/6 .

For x_> -3- , |x+3| = -(x+3) , so the expression becomes -1/(x-3) and the limit is
-1/-6 = 1/6 .

It's the two-sided limit x -> -3 that doesn't exist; the function has a jump discontinuity there and is undefined at x = -3.

For x-> +3 ,
the one-sided limits also exist, but they're infinite and don't match up.

|x+3| = 6 from either direction,

but lim x->+3- (x^2 - 9) approaches zero from negative values, so
the limit of the original expression goes to minus infinity
and lim x->+3+ (x^2 - 9) approaches zero from positive values, so
the limit of the original expression goes to positive infinity

So x-> +3 also does not exist, but for a different reason from the situation for x-> -3 .

(Sorry, I replaced my earlier post because I was getting confused about the actual problem under discussion.)

Last edited: Sep 30, 2007
8. Sep 30, 2007

### dynamicsolo

In general, it's not safe to say that a limit doesn't exist just because the denominator goes to zero in the limit. You need to look more carefully at the function. A simple example is
lim x-> 0 (1/x^2): the one-sided limits around zero are both positive infinity, so the two-sided limit is also.

Later in your course, you will learn about limits which are indeterminate ratios ( 0/0 or infinity/infinity ), which can turn out to be anything, depending upon the function involved; you will find out how to deal with them when you talk about L'Hopital's (Bernoulli's) Rule...

9. Sep 30, 2007

### Destrio

thanks for all the help everyone
I was initially able to solve the question which wanted the limit at -3
but I misread it as +3 which caused me the confusion with the undefined limit