limit doesn't exist

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February 14th 2006, 04:38 PM
asdfmaster

limit doesn't exist

lim[x->2] (x^2 - x + 6)/(x - 2)

I'm not sure the best way to solve this... but here's what I did:
after doing polynomial division, I got:

lim[x->2] (x+1+ 8/(x-2))

So, by reasoning, I see that the limit as x approaches 2 from the left does not equal the limit as x approaches 2 from the right (negative and positive infinity). Therefore, this limit does not exist.

My solutions manual says that the limit "does not exist since x - 2 -> 0 but x^2 - x + 6 -> 8 as x -> 2."

Someone please tell me what my book's explanation means.
February 14th 2006, 06:17 PM
MathGuru

n/0 dne for n not equal to zero

The only number that can be divided by zero to give another number is zero.

Therefore in order to show that the limit does not exist you can show that the limit approaches some non zero in the numerator and zero in the denominator. In this case:

8/0

does not exist.
February 14th 2006, 08:44 PM
CaptainBlack

Quote:

Originally Posted by asdfmaster

lim[x->2] (x^2 - x + 6)/(x - 2)

I'm not sure the best way to solve this... but here's what I did:
after doing polynomial division, I got:

lim[x->2] (x+1+ 8/(x-2))

So, by reasoning, I see that the limit as x approaches 2 from the left does not equal the limit as x approaches 2 from the right (negative and positive infinity). Therefore, this limit does not exist.

Would this form of argument then tell us that

$\lim_{x \rightarrow 2} \frac{x^2 - x + 6}{(x - 2)^2}$

exists and equals $\infty$ ?

RonL
February 14th 2006, 09:04 PM
asdfmaster

Thanks MathGuru & CaptainBlack

CaptainBlack, are you saying my reasoning is wrong?

if I graph
$<br /> \lim_{x \rightarrow 2} \frac{x^2 - x + 6}{(x - 2)^2}<br />$
I see that as x approaches 2 from the left and from the right, it increases without bound. I would then conclude that
$<br /> \lim_{x \rightarrow 2} \frac{x^2 - x + 6}{(x - 2)^2}=\infty<br />$
Also, if this reasoning is correct, then what MathGuru said:

Quote:

Therefore in order to show that the limit does not exist you can show that the limit approaches some non zero in the numerator and zero in the denominator.

would not be correct all the time...

I just started learning calculus! I'm sorry if these questions are very basic!
February 15th 2006, 03:56 AM
TD!

The limit doesn't exist because the upper limit doesn't equal the lower limit, i.e. you get a different result depending on whether you approach 2 from the right or from the left.

In the real numbers, we do not use the unsigned infinity in this context (this is done in complex analysis though), there is only a positive infinity and a negative infinity. Therefore you may not conclude that the limit "exists" and is equal to $\infty$ , that's wrong.
February 15th 2006, 04:39 AM
CaptainBlack

Quote:

Originally Posted by asdfmaster

Thanks MathGuru & CaptainBlack

CaptainBlack, are you saying my reasoning is wrong?

Neither the limit from the right or left exist.

RonL
February 15th 2006, 07:20 AM
MathGuru

Captain black was being a friendly skeptic earlier. He was trying to show you that sometimes the limit approaches the same from the left and from the right however the limit still does not exist.

My analysis that explains using numerator and denominator is more precise and also the same explanation your book was trying to give.
February 15th 2006, 11:09 AM
asdfmaster

thanks everyone!

Quote:

Captain black was being a friendly skeptic earlier.

That's what I assumed...but wasn't totally sure.

So what everyone is saying is that when a function appears to approach infinity, the limit does not exist???
So whenever you say the limit of a function is infinity, you are actually saying that the limit does not exist, but a convenient way to express that the limit does not exist is to write $\lim_{x \rightarrow a} f(x) = \infty$ ??

Quote:

Therefore in order to show that the limit does not exist you can show that the limit approaches some non zero in the numerator and zero in the denominator.

I'm just wondering, in every instance of finding a limit that exists, if the demoninator will approach zero, will numerator always approach zero too?
February 15th 2006, 01:28 PM
topsquark

Quote:

Originally Posted by asdfmaster

thanks everyone!
That's what I assumed...but wasn't totally sure.

So what everyone is saying is that when a function appears to approach infinity, the limit does not exist???
So whenever you say the limit of a function is infinity, you are actually saying that the limit does not exist, but a convenient way to express that the limit does not exist is to write $\lim_{x \rightarrow a} f(x) = \infty$ ??

I'm just wondering, in every instance of finding a limit that exists, if the demoninator will approach zero, will numerator always approach zero too?

For a limit to exist it has to approach a real number. Since infinity is not a real number, any limit that "blows up" (either in the positive or negative sense) does not exist.

Yes, the ONLY way for a limit to exist when the denominator of a rational expression approaches zero is for the numerator to approach zero as well. In this instance we can then apply L'Hospital's rule. However, be warned: Just because the numerator also goes to zero does NOT mean that the limit will exist. A trivial example of this: $\lim_{x \rightarrow 0} \frac{x}{x^2}$ does not exist.

-Dan