## Proving Limits don't exist

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Danikar
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### Proving Limits don't exist

I am taking a calculus 1 class and I am having trouble understanding a concept dealing with limits.

Take this problem
$\lim_{x\rightarrow2}\frac{x^2+x-6}{x-2}$
I can find that the limit is 5 by factoring the top and canceling the (x - 2). I have no idea how to work that out with limit laws though. My understanding of limit laws would make the denominator 0 and the everything is screwed at that point right?

Basically, I solved that problem, using the method above, and the very next problem was this.
$\lim_{x\rightarrow2}\frac{x^2-x+6}{x-2}$
This limit does not exist. I assumed it did not exist because I can do any factoring or replacing to make it look pretty. But I have no idea how to prove that it doesn't exist.

Any help? Did I explain this well at all?

EDIT: And per the rules, this is homework. But it is 2 problems of like 50.
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mark999
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### Re: Proving Limits don't exist

For the second function, what does it do near x=2? If you can demonstrate that it goes to infinity or minus infinity as x approaches 2 from above or from below, then you have proven that the limit doesn't exist at x=2.

Danikar
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### Re: Proving Limits don't exist

Yeah, I see how it is doing it. Because the numerator is always positive, so if the bottom is just a little below 2 the entire fraction is negative. And as it gets closer to 2 it get infinitely negative and if it is just a little bigger it stay positive and gets infinitely positive.

Is there a way to show this algebraically though? Or not so much?
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mark999
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### Re: Proving Limits don't exist

Do you know the "epsilon-delta" definition of a limit? You can show that for every positive number M, there is a number d such that the function is greater than M for all x between 2 and 2+d.

As x approaches 2 from above, the function diverges to infinity. The quadratic in the numerator has a minimum value, call it c. So the numerator is always at least c (and c is positive). If you take x between 2 and 2.1 then the function is always greater than c/(2.1-2), or 10c. If you take x between 2 and 2.01 then the function is always greater than 100c, and so on.

Danikar
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### Re: Proving Limits don't exist

Yeah, I have seen the epsilon-delta thing. I pretty much understand it now. Thanks for your help.
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doogly
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### Re: Proving Limits don't exist

For the first one, have you seen L'Hopital's rule?
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Danikar
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### Re: Proving Limits don't exist

I have seen it on a couple forums and what not. I really didn't understand it, but I am assuming that will probably be revealed to me soon.
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Talith
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### Re: Proving Limits don't exist

Do you know how to differentiate a function?

jestingrabbit
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### Re: Proving Limits don't exist

Neither of these need l'hopital's, nor the epsilon delta definition. A result you've probably used is that if [imath]\lim_{x\to a} f(x) = F[/imath] and [imath]\lim_{x\to a} g(x) = G[/imath] then [imath]\lim_{x\to a} f(x)g(x) = FG.[/imath]

Using this on the first one, note that $\lim_{x\to 2} \frac{x-2}{x-2} = 1$ and [imath]\lim_{x\to 2} x+3 = 5,[/imath] so

\begin{align*} \lim_{x\rightarrow2}\frac{x^2+x-6}{x-2} &= \lim_{x\to 2} \left(\frac{x-2}{x-2}\right)\left(x+3\right)\\ &= \left(\lim_{x\to 2} \frac{x-2}{x-2}\right) \left(\lim_{x\to 2}x+3\right)\\ &= 1* 5 = 5\end{align*}

So that was pretty easy.

The next one needs something a little bit trickier, namely: if [imath]\lim_{x\to a} f(x) = F\neq 0[/imath] then [imath]\lim_{x\to a} f(x)g(x)[/imath] exists if and only if [imath]\lim_{x\to a} g(x)[/imath] exists. Break the expression up into something whose limit at 2 you know, and something whose limit at 2 you know doesn't exist.
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Cleverbeans
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### Re: Proving Limits don't exist

Anything wrong with polynomial division? If we can show k + c/(x-2) goes to infinity for some constants k and c as x approaches 2 it should be pretty easy from there.
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gmalivuk
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### Re: Proving Limits don't exist

L'Hopital would be extra work for this. What factoring tells you is that, everywhere the denominator is not zero (in other words, everywhere you can divide by it and make sense), the function reduces something simpler. This means that while the rational function given has a hole where the denominator is zero, it behaves quite simply everywhere else. (It's much harder to prove L'Hopital's than to prove that two functions equal everywhere but one point have the same limit approaching that point.)
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doogly
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### Re: Proving Limits don't exist

L'Hopital may be extra artillery, but OP said "I can find that the limit is 5 by factoring the top and canceling the (x - 2). I have no idea how to work that out with limit laws though," and I figure if you are looking for a more general concept than the factoring that may be it. Factoring is perfectly adequate though, and it is consistent with the limit laws, as said. [Though for really big polynomials, it's faster to take their derivatives than factor them anyway.]
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gmalivuk
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### Re: Proving Limits don't exist

How to best work out how factoring works with limit laws is basically like I said, though: dividing by the denominator where it isn't zero tells you the value of the function everywhere else, and it's pretty easy to show that functions identical everywhere but a specific point or two must have the same limits everywhere, including those points.
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Danikar
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### Re: Proving Limits don't exist

Cool, from what I have read in here I came up with this. Which seems mostly correct. I use DNE for does not exist. Is there a better notation for that? Anyway, thanks for the help.

Original problem
$\lim_{x\to 2} \frac{x^2 - x + 6}{x - 2}$

Break down problem into seperate functions
\begin{align*}f(x) &= x^2 - x + 6\\ g(x) &= \frac{1}{x - 2}\\ \lim_{x\to 2} \frac{x^2 - x + 6}{x - 2} &= \lim_{x\to 2} f(x)g(x)\end{align*}

Get limit for f(x)
$\lim_{x\to 2} f(x) = 8$

jestingrabbit wrote:if [imath]\lim_{x\to a} f(x) = F\neq 0[/imath] then [imath]\lim_{x\to a} f(x)g(x)[/imath] exists if and only if [imath]\lim_{x\to a} g(x)[/imath] exists

\begin{align*}\lim_{x\to 2} g(x) &= DNE \Rightarrow\\ \lim_{x\to 2} f(x)g(x) &= DNE \Rightarrow\\ \lim_{x\to 2} \frac{x^2 - x + 6}{x - 2} &= DNE\end{align*}
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jestingrabbit
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### Re: Proving Limits don't exist

Looks fine to me. I thought cleverbeans' suggestion was pretty good too.

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Yesila
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### Re: Proving Limits don't exist

Just a general warning for the OP. Regardless of the truth of theorems posted here if you don't have the theorem either presented in your class or sitting in a section of your book that you have already covered you should either avoid using them or be prepared to prove them.

jestingrabbit
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### Re: Proving Limits don't exist

Yesila wrote:stuff

These are good points.

Regarding the proof, if you look at the theorem that states approximately "the product of the limits is the limit of the products" ie the one I used for the first problem, then you can start to see where the second result I stated comes from. Label statements as follows:

A) [imath]\lim_{x\to a} f(x)=l[/imath];
B) [imath]\lim_{x\to a} g(x)=L[/imath]; and
C) [imath]\lim_{x\to a} f(x)g(x)=lL[/imath].

The theorem I used to work the first result can be rewritten as A and B imply C. If you can show that A and C imply B, then you're done, but it turns out that's not true. For instance, if a=0, f(x)=x2 and g(x)=1/x, then you can see that's a counter example. So, what is it that assuming l=/=0 gives us? Hint:
Spoiler:
[imath]\lim_{x\to a} \frac{1}{f(x)}=\frac{1}{l}.[/imath]

If you don't even have the first result in your notes then its pretty clear that the purpose of the exercises is to practise applying the definition. Your teacher would have to be a bastard to set that sort of task, but there are bastards out there.
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Danikar
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### Re: Proving Limits don't exist

\begin{align*} \lim_{x\to 2} \frac{x^2-x+6}{x-2} &= \frac{2^2 - 2 + 6}{0}\\ &= \frac{8}{0} \end{align*}

My professor said just to say that any constant over 0 means that the limit is undefined. and [imath]\frac{0}{0}[/imath] = more work.

Was mainly looking to comprehend that concept a little more.
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Why Two Kay
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### Re: Proving Limits don't exist

When you learn L'Hopital's rule you will be able to evaluate the indeterminate cases such as 0/0 and infinity/infinity. The reason for the need to do this is because while both the numerator and denominator are going to zero as x->whatever, they may be going at different rates, which could mean the actual limit is 0, 1, 17, 958, etc, some actual value.

In general, when you evaluate a limit by "plugging in":

constant/zero --> Diverges/To Infinity/To Negative Infinity
constant/infinity --> Zero

zero/constant --> Zero
zero/zero --> Indeterminate, find some other way.
zero/infinity --> Zero

infinity/constant --> Diverges/To Infinity/To Negative Infinity
infinity/zero --> Diverges/To Infinity/To Negative Infinity
infinity/infinity --> Indeterminate, find some other way

In these Indeterminate cases, you can use L'Hopital's rule, which you probably have not covered yet, to solve them much easier than doing some other strange things they tried to teach us. Our limits unit in AP Calc BC was in August/September. We learned L'Hopital's rule last Wednesday. I can't figure out why they took that long to teach us (they taught us the day before improper integrals too).
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btilly
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### Re: Proving Limits don't exist

One reason to teach L'Hopital's rule late is that you can't use it to evaluate limits until you understand derivatives, and you won't understand derivatives until you understand limits. Therefore if you want people to understand limits OR derivatives, you have to teach limits first, without L'Hopital's rule. Which ties to the often ignored point that what you are taught in math is mostly about helping you understand math, and not about getting the right answer.

Another reason to discourage the premature use of L'Hopital's rule is that it only works if there is an indeterminate form. I have personally seen people learn L'Hopital's rule, forget that condition, and then try to use it on limits that aren't an indeterminate form, thereby getting lots of wrong answers.

Finally a random point. If you have some basic common sense about relative orders of magnitude, all standard textbook problems involving L'Hopital's rule are readily solvable without it. (There are some pathological cases where it solves problems that are nearly impossible without it, but they do not appear as standard textbook problems.) In the long run that common sense will serve you a lot better than a magic formula. Therefore I for one would not consider it a significant loss if the rule was dropped entirely from the course rather than just being left for a long time.
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### Re: Proving Limits don't exist

btilly wrote:Finally a random point. If you have some basic common sense about relative orders of magnitude, all standard textbook problems involving L'Hopital's rule are readily solvable without it. (There are some pathological cases where it solves problems that are nearly impossible without it, but they do not appear as standard textbook problems.) In the long run that common sense will serve you a lot better than a magic formula. Therefore I for one would not consider it a significant loss if the rule was dropped entirely from the course rather than just being left for a long time.

I agree that many people are generally way too quick to invoke l'Hopital (like in this thread, or evaluating the limit at 0 of sin(x)/x), but I still think it's the best technique often enough to justify learning it. I'm thinking of problems like this:$\lim_{x\to\infty}(2x-1)\ln\frac{x+1}{x-2}$Sure, you could use polynomial long division, linearization, and order-of-magnitude simplification to get the same answer as l'Hopital, but it's not what I'd call a "ready" solution and certainly not what I would call common sense.

btilly
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### Re: Proving Limits don't exist

Cosmologicon wrote:
btilly wrote:Finally a random point. If you have some basic common sense about relative orders of magnitude, all standard textbook problems involving L'Hopital's rule are readily solvable without it. (There are some pathological cases where it solves problems that are nearly impossible without it, but they do not appear as standard textbook problems.) In the long run that common sense will serve you a lot better than a magic formula. Therefore I for one would not consider it a significant loss if the rule was dropped entirely from the course rather than just being left for a long time.

I agree that many people are generally way too quick to invoke l'Hopital (like in this thread, or evaluating the limit at 0 of sin(x)/x), but I still think it's the best technique often enough to justify learning it. I'm thinking of problems like this:$\lim_{x\to\infty}(2x-1)\ln\frac{x+1}{x-2}$Sure, you could use polynomial long division, linearization, and order-of-magnitude simplification to get the same answer as l'Hopital, but it's not what I'd call a "ready" solution and certainly not what I would call common sense.

Personally I'd go the other way on that problem. It is easier for me to recognize that $\ln\frac{x+1}{x-2} = \ln(1 + \frac{3}{x-2}) \approx \frac{3}{x-2}$ from which the answer to the original question is obviously 6 than it is to differentiate that expression. Furthermore it comes "hidden" as a candidate for the standard version of L'Hopital's rule because it is not written with a division and you have to do some algebraic manipulation to get it into that form. (Which manipulation makes the derivative even uglier.)
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### Re: Proving Limits don't exist

I don't think it's hidden. 0 * infinity is one of the standard indeterminate forms that l'Hopital is good for. If you don't know what to do with it, I would say you don't know how to use l'Hopital's rule. The derivative is kind of messy, yes, but each step in evaluating this limit is straightforward with l'Hopital. It doesn't require any "lateral thinking". Your method also works, and it's no messier, but I think it's much less obvious, not more obvious. YMMV.

But if you like, I'm sure you can easily imagine a similar limit where the indeterminate form is already a quotient. EDIT: like this:$\lim_{x\to 0}\frac{\tan{x}-x}{x-\sin{x}}$I'm sure it can be done your way, but come on. Doesn't it just scream l'Hopital?