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I had a question about finding limits of piecewise functions through graphs. I believe, I am missing something in my fundamentals about finding limits for these functions. Firstly, I would like to confirm that the empty dot represents a "hole" and the point is not included in the function of the line. For example, line closest to $h(x)$ is an example. Therefore, a coloured in dot represents a point that is on the graph e.g. $(3,2)$. So now for my main question

How do I determine the limit of a $x$ that has two $y$'s?

For example, the $\lim_{x\to 3^+} h(x)$ The back of the book says the answer is 3 but how can that be the case when, to my knowledge, an empty dot means that the point is hole? Or if that is not the case, then why is the answer not 2?

If someone could please clear this up for me, that would be great.

Thanks for the help!

P.S I just decided to add the questions in there for reference, I am not asking for anyone to solve them and please don't. I prefer that you give me the tools to do it myself or some hints.

A piecewise function that has 3 different conditions

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Each $x$ on the domain has only one image (for example, $f(-1)=1$). So you have to check lateral limits. Don't worry with two circles. –  Sigur Feb 4 '13 at 22:31
    
The notation $x\to a^+$ means that we are approaching $a$ from the side where $x>a$ (in crude terms, the RHS), similarly $x\to a^-$ means that we are approaching it from the LHS. The fact that the dot is empty does not mean that the limit doesn't exist: $\lim_{x\to a}$ is just not necessarily equal to $f(a)$ –  L. F. Feb 4 '13 at 22:33
    
When finding the limit in question, you are considering values of $x$ strictly larger than $3$ but close to $3$ (such as $3.000001$). What is, approximately, the value of $h(x)$ for such $x$? –  David Mitra Feb 4 '13 at 22:33
    
@L.F. Yes, I know that if the empty dot is there then it is is possible to have limit but if the empty dot means that the other one is the limit then why is the answer not 2? –  gekkostate Feb 4 '13 at 22:35
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@gekkostate I don't understand what you're doing - it is crystal clear that if you put a pencil on the curve, approach $x=3$ from the RHS without lifting the pencil, you arrive at a point where $y=3$ not $2$. –  L. F. Feb 4 '13 at 22:45
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2 Answers

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When evaluating limits, it doesn't matter whether the dot is filled in or not. Intuitively, the following two questions are the same:

What (if anything) is $\lim_{x\to 3^+}h(x)$?

If you were to approach the vertical line $x=3$ along the curve $y=h(x)$ from the right, what $y$-value (if any) would you approach?

Considered in the latter light, imagine tracing the path of the curve $y=h(x)$ and getting closer and closer to the line $x=3$. If we want to get to within a certain distance of the line $y=3$, then all we have to do is get close enough to the line $x=3$ from the right (this should be clear from the graph). That is what it means to say that $\lim_{x\to 3^+}h(x)=3$. We cannot do the same thing with the line $y=2$, since no matter how close to the line $x=3$ we get on the right, we keep tracing over points that are more than $1$ unit away from $y=2$.

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Okay, I think I understand what you are getting at. So suppose then I approach $3^-$ then the answer will be 2 right? Because we are now coming from the LHS which is represented by the linear line. –  gekkostate Feb 4 '13 at 22:46
    
Exactly. Note that the answer from the LHS would still be $2$ even if that dot were also unfilled. –  Cameron Buie Feb 4 '13 at 22:47
    
Great! I understand the problem I was having now. I was confusing the tracing bit. Instead of following the function, I was following the x-axis from right to left and vice versa. Thanks a bunch! :D –  gekkostate Feb 4 '13 at 22:49
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When you say that $\lim_{x\to 3^+}h(x)=3$ it is not necessary to have $h(3)=3$.

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Yes, I understand that but why is the value not 2? That's the main problem that I am having. –  gekkostate Feb 4 '13 at 22:37
    
You can consider a small open interval around $2$ (for example, $I=(1.9,2.1)$) such that it is not possible to find $x>3$ with $f(x)\in I$. So you can isolate $2$ from the image $f((3,\infty))$. –  Sigur Feb 4 '13 at 22:40
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