Need a little guidance - Limits

[Barchie]

Homework Statement

For f(x) = |x^2-9|/x-3 and a=3, discuss the limiting behavior of f(x) as x->a+, as x->a- and as x->a

The Attempt at a Solution

Am i right to say that the limit for x-> 3 is undefined? I am also a little confused with the a+ and the a-

for a+ i would say that 3+^2-9/3+-3 = 0+/0+ = 1 and for a- it would = -1 ... but i get the feeling that this is wrong... any guidance?

Homework Statement

For f(x) = |x^2-9|/x-3 and a=3, discuss the limiting behavior of f(x) as x->a+, as x->a- and as x->a
EDIT: ok so putting the values into a calculator give me 6 & -6 i am content with those answers... i am also think i am getting closer with the idea that i can factor the top line into:

(x+3)(x-3) and then cancel the (x-3)'s leaving me with x+3 which would explain 6, but not -6.

can i cancel the denominator with the abs value numerator?

Thanks for your time.

 

[Barchie]

ok no worries. have resolved this issue.

for absolute value |x^2-9| i factor to |(x+3)(x-3)|

for x < 3 | -(x+3)(x-3) which evaluates for 3- to:

(-x-3)(x-3)/x-3 | cancels to:
-x-3
lim x-> 3- = -6

for the 3+

(x+3)(x-3)/(x-3) cancels to:
x+3
lim x-> 3+ = 6

and if the limits from the right hand side and the left don't equal then the limit is undefined.

Hope this helps someone, and if not... sorry for wasting your time.

 

[praharmitra]

Your answer is exactly right. To explain the -6, why don't you write the x->a- limit as follows

x = a-h and h->0, keeping in mind that h>0 (this would help you in working with the absolute values. Try it.

 

[MednataMiza]

Hi!
I have a few questions:
Those a+ and a- means that you are looking at a right and a left limit ?
Second How do you get +1 from 0+/0+ ? That is an indeterminate form.
Now: Using the l'Hôpital's rule you get 2x and this is the answer of your question, why you get +6 & -6 on the calculator.
Yes,you can cancel it,but first you have to determine how to 'open' this abs value brackets and then depending on this you can cancel them :)
Edit: Sorry praharmitra didn't see your post ...

 

[praharmitra]

Hi!
I have a few questions:
Thos a+ and a- means that you are looking at a right and a left limit ?
Second How do you get +1 from 0+/0+ ? That is an indeterminate form.
Now: Using the l'Hôpital's rule you get 2x and this is the answer of your question, why you get +6 & -6 on the calculator.
Yes,you can cancel it,but first you have to determine how to 'open' this abs value brackets and then depending on this you can cancel them :)
Edit: Sorry praharmitra didn't see your post ...

1. a+ and a- do mean the right and left limits.

2. 0+/0+ is not equal to +1. It has been wrongly written. What you actually get is something of the form
<br /> Lt_{h\rightarrow0} \frac{h}{h} = 1<br />
that is what has been written in shorthand above.

3. You have incorrectly used L'hopital rule.
<br /> \frac{d(|x|)}{dx} = 1, \ x&gt;0;\ -1,\ x&lt;0<br />

 

[MednataMiza]

Yes,I have used the l'Hôpital's rule right.That's just the basic form :)