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I'm doing some homework for a calculus course I am taking and, though I feel silly because I've done problems like this before, I'm having some issues. I feel like I am getting the wrong answers for a couple of problems, and I'd like a second (or third, or tenth) opinion.

1) Find $\lim_{x\to 1}f(x)$ where: $$f(x) = \left\{ \begin{array}{cc} x^2+2, & x\neq 1\\ 1, & x=1 \end{array} \right.$$

I'm thinking that the limit does not exist, since you have to plug 1 in to the equation, making x=1, meaning f(x) =1. Please, correct me if I'm wrong.

2) $\lim_{x\to -1} \frac{x^3-1}{x+1}$

I said it is undefined, but I'm not sure...

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Thanks for fixing that for me... I was trying to figure out how to do it and I have no idea. :x – Lish May 20 '11 at 9:08
Learn LaTeX – user9413 May 20 '11 at 9:11
You really have to tell us your definition of limit. There are (at least) two conventions, that include or exclude the actual limit value. Judging from your question, I would suspect that your definition excludes the value at 1, but you have to look up definitions to be able to do your homework. – Phira May 20 '11 at 9:16
up vote 3 down vote accepted

For the $2$ problem the as $x \to -1$ $f(x) = \displaystyle\frac{x^{3}-1}{x+1} \to -\infty$. Here is the graph

enter image description here

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There are some signs missing. It's $f(x) \to -\infty$ as $x \to -1^+$. But you get $f(x) \to +\infty$ as $x \to -1^-$ (that part is missing from your graph), so the limit doesn't exist even in the improper sense (unless you're working in the extended complex numbers, where there's only one $\infty$). – Hans Lundmark May 20 '11 at 10:32
@Hans: Thanks a lot Hans. – user9413 May 20 '11 at 10:46

Limits are not about "plugging in" a value of $x$. The closest explanation of how to think of a limit would be to "plug in" values of $x$ that are close to the limiting value of $x$. So in the first exercise, you should consider $x$ very close to $1$, and notice that values very close to $3$ come out. Do something similar for the second limit after you factor the numerator as $(x+1)(x^2-x+1)$.

If you simply let $x$ be the value $1$ or $-1$ in these problems, then you do not yet really understand what it means to take a limit.

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What you've factorized is $x^3+1$, but in the question it actually says $x^3-1$. – Hans Lundmark May 20 '11 at 10:34
@Hans: Yes, that is the mistake he has made. – user9413 May 20 '11 at 10:48
Ah yes, I misread the question. – alex.jordan May 21 '11 at 3:17
Still, my point remains. Even using a graph is an unreliable method for evaluating a limit. For the second limit, one should consider $x=-1\pm\epsilon$, a number slightly removed from $-1$. Then $f(x)=\frac{-2\pm 3\epsilon-3\epsilon^2\pm\epsilon^3}{\pm\epsilon}=\mp\frac{2}{\epsilon}+3\mp 3\epsilon+\epsilon^2$, which is more clearly $-\infty$ when you started with $+\epsilon$ and $\infty$ when you started with $-\epsilon$. – alex.jordan May 21 '11 at 3:50

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