# Limits of functions

1. Oct 8, 2012

### Artusartos

Let S be a subset of $R$, let a be a real number or symbol $+ \inf$ or $- \inf$ that is the limit of some sequence in S, and let L be a real number or symbol $\inf$ or $- \inf$. We write $lim_{x \rightarrow a^s} f(x) =L$ if

1) f is a function defined on S,
2) for every sequence $(x_n)$ in S with limit a, we have $lim_{n \rightarrow \inf} f(x_n) = L$

a) For $a \in R$ and a funciton f we write $lim_{x \rightarrow a} f(x) =L$ provided $lim_{x \rightarrow a^s} f(x) = L$ for some set S = J \ {a} where J is an open interval containing a. $lim_{x \rightarrow a} f(x) =L$ is called the [two-sided] limit of f at a. Note that f need not be defined at a and, even if f is defined at a, the value f(a) need not equal $lim_{x \rightarrow a} f(x) = L$. In fact, $f(a) = lim_{x \rightarrow a} f(x)$ if and only if f is defined on an open interval containing a and f is continuous at a.

Can anybody please explain this to me...it just seems very confusing...

a) Why can't {a} be contianed in S? (When it says "$lim_{x \rightarrow a^s} f(x) = L$ for some set S = J \ {a} where J is an open interval containing a")?

b) Also, this seems kind of strange to me...

First it says "Note that f need not be defined at a and, even if f is defined at a, the value f(a) need not equal $lim_{x \rightarrow a} f(x) = L$"...and then it says "$f(a) = lim_{x \rightarrow a} f(x)$ if and only if f is defined on an open interval containing a and f is continuous at a.".

So if f is defined on an open interval containg a, then it must also be defined at a, right? Then why does it say that f doesnt need to be defiend at a? Also, if f is continuous at a ,shouldn't it be defined there too? Finally, why does the value f(a) not need to equal $lim_{x \rightarrow a} f(x) = L$?

2. Oct 8, 2012

### MarneMath

I'll focus on this statement, and allow someone with a deeper understanding of limits than I handle the more technical aspect of the definition.

Consider the function (x^2-1)/(x-1), let our interval be the set of Real numbers, then clearly in our interval the function is not defined at x = 1. However, we can rewrite this function as ((x-1)(x+1)/(x-1) = x+ 1. This function is the same except at x = 1, where it is defined. Turns out if we take the limit of the original function at x = 1, and use the fact that the original function is the same as x + 1, then it becomes clear to us that the limit of the original function is 2 as x approaches 1.

That's how you can have a function defined on an interval with a function that is not defined, but still have a limit that exist.

3. Oct 8, 2012

### micromass

The answer to your problems is that if we look at

$$\lim_{x\rightarrow a} f(x)=L$$

then we simply don't care about the value of f at a.

Let's first look at the notion of continuity. We can very easy describe what we intuitively mean with a continuous function: it is a function that we can draw "without lifting our pen from the paper". For example

is not continuous in 0 because we have to lift our pen up when we draw it.

Now, this definition of continuity is of course not mathematically rigorous. A better definition would be: f is continuous in a if for all x "close to" a, we have that f(x) is "close to" f(a). So we send close points to close points. This is not rigorous, because we have not said what we meant with "close to", but it will suffice for now.

Now, not all functions are continuous of course. For example, look at

This function is not continuous in the point a=2. However, IF we were to redefine the function f in 2 as f(2)=1, then the function would be a nice, continuous function. We say that $\lim_{x\rightarrow 2} f(x)=1$. So we say that $\lim_{x\rightarrow a} f(x)=L$ if the function would be continuous in a if we only had defined f(a)=L.

So, if we want to find the limit at a, then we care about how we would (re)define the function f such that it would be continuous. The current value of f at a doesn't matter. For example, we may have f(a)=1000 or f(a)=10000, we don't care. We might even have that f is undefined at a, that doesn't matter for the calculation of the limit.

I hope this makes some things clear.

4. Oct 8, 2012

### Artusartos

Thanks, but can you tell me what S and J are in this case?

5. Oct 8, 2012

### Artusartos

Thanks a lot. Can you also tell me what S and J are in the example that you gave?

6. Oct 8, 2012

### HallsofIvy

J, just as you say, can be any open interval of real numbers that contains a. S is just J with a itself removed (sometimes called a "deleted interval"). In the last example Artusartos gave, J could be, possibly, the open interval (3/2, 5/2) and S would be (3/2, 2) union (2, 5/2).

7. Oct 8, 2012

### Artusartos

So let's say we have this...

$lim_{x \rightarrow 2} \frac{x^2 - 4}{x-2}$...

Our textbook says that we can rewrite this as $lim_{x \rightarrow 2} x+2$

So, in this case what exactly is $lim_{x \rightarrow a^s} f(x) = L$ and $lim_{x \rightarrow a} f(x) = L$?

In other words, can you restate " For $a \in R$ and a funciton f we write $lim_{x \rightarrow a} f(x) =L$ provided $lim_{x \rightarrow a^s} f(x) = L$ for some set S = J \ {a} where J is an open interval containing a.", using the example that I gave?