one more limit question i need help with.

[renegade05]

\(\displaystyle \lim_{x\to3\textsuperscript{+}} \frac{\left|x^2-7x+12\right|}{9-x^2}\)

I factored the numerator to \(\displaystyle (x-3)(x-4)\) but now i am stuck. I dont even know if thats the way to go about it.

Need some direction.
Thanks.

 

[Loren]

Does it help any to realize that 9-x[sup:3hg4jhji]2[/sup:3hg4jhji] factors into (3-x)(3+x) which can be rewritten as -(x-3)(x+3)?

 

[renegade05]

Does it help any to realize that 9-x[sup:3q2x2gaj]2[/sup:3q2x2gaj] factors into (3-x)(3+x) which can be rewritten as -(x-3)(x+3)?

difference of squares..of course... no i did not see that!! Thanks.

I think i can take it from here, are the absolute value signs gonna play a role here? They are intimidating me...

 

[DrSteve]

They do play a role. Remember that you're taking a limit from the right. So you are looking at x values greater than 3, but VERY close to 3 (in particular, less than 4). To get a handle on this, try plugging in values like 3.01, 3.001, etc. for x and see what happens (and also think about why this is happening). Also think about what would happen if the absolute value wasn't there, and what happens if you take the limit from the left.

 

[renegade05]

They do play a role. Remember that you're taking a limit from the right. So you are looking at x values greater than 3, but VERY close to 3 (in particular, less than 4). To get a handle on this, try plugging in values like 3.01, 3.001, etc. for x and see what happens (and also think about why this is happening). Also think about what would happen if the absolute value wasn't there, and what happens if you take the limit from the left.

Ok. I understand all or most of this. I know the answer to be -1/6. My question now is: why is there no asymptotic behavior at x = 3. Since 9-x^2 does factor to -(x-3)(x+3) should there not be asymptotes at x=3 and x=-3 so shouldnt the limit approach -infinity?

 

[DrSteve]

They do play a role. Remember that you're taking a limit from the right. So you are looking at x values greater than 3, but VERY close to 3 (in particular, less than 4). To get a handle on this, try plugging in values like 3.01, 3.001, etc. for x and see what happens (and also think about why this is happening). Also think about what would happen if the absolute value wasn't there, and what happens if you take the limit from the left.

Ok. I understand all or most of this. I know the answer to be -1/6. My question now is: why is there no asymptotic behavior at x = 3. Since 9-x^2 does factor to -(x-3)(x+3) should there not be asymptotes at x=3 and x=-3 so shouldnt the limit approach -infinity?

There is something very fundamental here that you're confused about. You seem to be thinking about the theorem which says that if substituting in \(\displaystyle c\) for \(\displaystyle x\) into a rational function makes the denominator zero, but the numerator nonzero, then there is a vertical asymptote of \(\displaystyle x=c\). This theorem cannot be applied here, because substituting in \(\displaystyle 3\) for \(\displaystyle x\) makes both the numerator and denominator zero. This is called an indeterminate form. This means that more work is required to determine if there is an asymptote there. What kind of work? Well you need to compute the limit using the methods you've learned in calculus.

 

[renegade05]

Are you saying that if it is not 0/0 say any real over 0 there will be an asymptote? And if it is 0/0 it is indeterminate and there may or may not have an asymptote?

How would I differentiate finding the limit of 3- and 3+ of this equation analytically ?? No graphing...

 

[DrSteve]

Are you saying that if it is not 0/0 say any real over 0 there will be an asymptote? And if it is 0/0 it is indeterminate and there may or may not have an asymptote?

Yep - that is correct.

I think at this point I can go ahead and give you a complete algebraic solution:

\(\displaystyle \lim_{x\rightarrow 3^+}\frac{|x^2-7x+12|}{9-x^2}=\lim_{x\rightarrow 3^+}\frac{|x-3||x-4|}{(3-x)(3+x)}=\lim_{x\rightarrow 3^+}\frac{(x-3)(4-x)}{(3-x)(3+x)}=\lim_{x\rightarrow 3^+}\frac{(-1)(4-x)}{(3+x)}=-\frac{1}{6}\)

\(\displaystyle \lim_{x\rightarrow 3^-}\frac{|x^2-7x+12|}{9-x^2}=\lim_{x\rightarrow 3^-}\frac{|x-3||x-4|}{(3-x)(3+x)}=\lim_{x\rightarrow 3^-}\frac{(3-x)(4-x)}{(3-x)(3+x)}=\lim_{x\rightarrow 3^-}\frac{(4-x)}{(3+x)}=\frac{1}{6}\)


The main point here is that \(\displaystyle |x-3|\) is \(\displaystyle x-3\) if \(\displaystyle x\) is a little bigger than \(\displaystyle 3\), and \(\displaystyle |x-3|\) is \(\displaystyle 3-x\) if \(\displaystyle x\) is a little smaller than \(\displaystyle 3\).

Also note that in both cases \(\displaystyle x<4\) so that \(\displaystyle |x-4|\) is \(\displaystyle 4-x\).