# Going over a test: limits

I have a test to review, would it be possible to just take a picture of the pages and link them here or is that now allowed? If not I will just pick a few problems out of it.

I was trying to find the limit $$\lim_{x \to 0} \frac{x-1}{x^3 +2x^2} .$$ I was told I could find this logically on the test so I wrote down that the answer is - infinity stating that the degrees of the denominator are greater than the degrees in the numerator so it will go to infinity. This was wrong.

I was told to find the precise definition of the limit of $F(x)$ as $x$ approaches $\infty$ is equal to $8$. I said that as $x$ approaches $\infty$ the limit of $f(x)$ is $8$. $f(x)$ cannot equal $8$ but only approach is as close as you want it to. That of course was wrong.

I am just wondering why I am so bad at math, maybe someone can see a trend in here.

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I think the bit about $8$ is a totally separate question. –  Dylan Moreland Sep 19 '11 at 23:31
I edited it, it should make more sense now. –  user138246 Sep 19 '11 at 23:52
The main problem I see is that you don't use/know the definition of a limit. Obviously the professor will not give you a high grade if you say the definition of "The limit of $f(x)$ as $x$ approaches $\infty$ is $8$" is "As $x$ approaches $\infty$ the limit of $f(x)$ is $8$". –  TMM Sep 19 '11 at 23:58
@Thijs I am not seeing a difference at all, what did I have wrong? I know what a limit is, how is what I said wrong? –  user138246 Sep 20 '11 at 0:06
@Jordan: It is not wrong, but that does not mean it is the desired answer. If someone asks you to prove that $\zeta(2) = \pi^2/6$, then you could say "Obviously $\zeta(2) = \pi^2/6$" which is not wrong or different from what he said, but not what he wanted to hear either. Here the professor wants you to give a formal definition of a limit, which you didn't. –  TMM Sep 20 '11 at 0:10

Your reasoning in the first answer is incorrect. The degrees don't really have any direct bearing on the fact that the limit does not exist. You are probably getting confused with the criteria for what happens to a rational function as $x\to\infty$ (the "end behavior", or the "horizontal asymptotes"). But they have nothing to do with the issue here.

Instead, you need to try to understand what is happening with this limit. $$\lim_{x\to 0}\frac{x-1}{x^3+2x^2}.$$ We want to try to understand what happens to the values of $$\frac{x-1}{x^3+2x^2} = \frac{x-1}{x^2(x+2)}$$ as $x$ gets closer and closer to $0$. If $x$ is very close to $0$, the value of $x-1$ is going to be very close to $-1$; the value of the denominator is going to be very close to $0$ (because $x^2$ is going to be very close to $0$, $x+2$ is going to be very close to $2$, and so the product is going to be very close to $0$); in addition, the denominator is going to be positive, because both $x^2$ and $x+2$ will be positive.

So for $x$ very small, the fraction will be a number very close to $-1$ divided by a positive number very close to $0$. The result will be negative (negative number divided by positive number), and have very large size (dividing something very close to $-1$ by something very close to $0$ gives you something very large in size). As $x$ gets closer and closer to $0$, that denominator is getting closer and closer to $0$, and the quotient gets larger and larger and larger in size, while staying negative. And that is why the limit is $-\infty$. Not because of the degrees. The same would be true if you had $$\lim_{x\to 0}\frac{x^3-1}{x^2},$$ where the limit is also equal to $-\infty$ (by the same argument as above) even though the degree of the denominator is now smaller than the degree of the numerator. Which is why invoking the degrees is incorrect: they don't actually tell you anything useful here.

As to the second one: the "precise definition" cannot be to simply repeat the statement with the words in a different order. Encyclopedias and dictionaries would be rather silly if that were the case...

Instead, you are being asked to find the precise way in which the limit is defined. Your book almost certainly has a definition in a highlighted box somewhere, either with words or with epsilons ($\varepsilon$) and deltas ($\delta$). It is false that $f(x)$ cannot equal $8$, as I noted in a comment: the function $f(x)=8$ has limit $8$ as $x\to\infty$, and the value is always $8$. You are confusing the oft-repeated statement that when you consider a limit as $x$ approaches $a$, we never consider $x$ equal to $a$. But that's the variable, not the limit itself. Here, we are talking about the values of the function. If you have $$\lim_{x\to 5}f(x) = 8,$$ then we do not consider what happens when $x$ is $5$, only what happens when $x$ is near $5$. But nowhere do we prohibit $f(x)$ to take the value $8$ for $x$s close to $5$. Again, you seem to be misremembering something and applying it to what it is not about.

The "wordy" definition says something like: "we can make all the values of $f(x)$ be as close as we want to $8$ provided that $x$ is large enough", or similar words. The formal definition will say: "for every $\varepsilon\gt 0$ there is an $M\gt 0$ such that if $x\gt M$, then $|f(x)-8|\lt\varepsilon$." At least one of these two ideas should have been covered in your course at some point, at least for regular limits if not also for limits at infinity. It is also entirely possible that you are being asked to search for the precise definition somewhere else, like, say, a book in the library.

I'm going to guess, on the basis of your past questions, that much of your confusion arises in large part because you are trying to memorize everything, and then you get confused because you remember the wrong thing, or you remember it incorrectly. Let me warn you: you won't be able to pass a calculus course through sheer memorization prowess; it really does require some understanding.

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Thanks, you made me realize I have not actually learned anything at all in this class. I have no idea what is going on. I was so stupid that I actually thought I did, and I felt confident going into the test, but I ended up with a failing grade. Time to drop out of school I guess. –  user138246 Sep 20 '11 at 12:13