# How to discover algebraically when a function has no limit?

1. Aug 20, 2013

### CuriousBanker

I intuitively and graphically understand when a function has no limit...basically if the value it approaches from opposite directions don't agree. So if from the left side it approaches negative infinity and the right side it approaches positive infinity, does not exist...I get that. But algebraically, can you give me a couple of examples of functions with no limits, and how I would know that without looking at a graph?

2. Aug 20, 2013

### micromass

That's not the whole story. It is true that if the left-hand side and the right-hand side limit don't agree, then he limit doesn't exist. This is called a jump discontinuity. But it can also happen that the left-hand side and right-hand side limit don't exist. In that case, it also has no limit. A famous example is $f(x) = \sin(1/x)$ in $0$:

This is called an essential discontinuity. Those usually look very ugly like the above.

It's mostly experience. But maybe you can benefit from these examples:

Example 1:
Take the function
$$f(x) = \left\{\begin{array}{l} x~\text{if}~x<0\\ x+2~\text{if}~x\geq 0\end{array}\right.$$

This function is piecewise defined. That is, we give a separate definition of the function in the case $x<0$ and $x\geq 0$. In this case, you can suspect that the limit in $0$ might not exist. Of course, some piecewise defined functions are continuous, so it's not because something is piecewise defined that the limitss don't exist. You still need to check it formally by finding the left-hand side and right-hand side limits.

Another example is

$$f(x) = \left\{\begin{array}{l} \frac{\sin(x)}{x}~\text{if}~x\neq 0\\ 1000~\text{if}~x=0\end{array}\right.$$

This function is piecewise defined. So it makes sense to suspect the limit in $0$ doesn't exist. However, you can easily check that it in fact does exist.

Example 2:
When dealing with infinities, limits might not exist. For example, the function $f(x)=1/x$ has no limit in $0$ since the left-hand side and right-hand side limits don't equal.

Example 3:
When you get infinities inside a function, limits might not exist. For example, the limit of $f(x) = \sin(1/x)$ doesn't exist in $0$. A way to see this is to plug in $0$ to get $\sin(\infty)$. This makes no sense. So this is an indication that the limit might not exist. You still need to check it rigorous.

However, take $f(x) = x\sin(1/x)$. This limit does exist.

3. Aug 20, 2013

### CuriousBanker

f(x)=⎧⎩⎨sin(x)x if x≠01000 if x=0 How does the limit exist in this example? Doesn't seem like it does to me.

Also, why doesnt the limit of f(x)=sin(1/x) in 0 exist? I don't fully get that one.

The other ones I get, thanks. I just didnt know if there was a formal way of proving a limit doesnt exist or if you just do it by looking at it

4. Aug 20, 2013

### HallsofIvy

You seem to have too many left braces in that! Do you mean f(x)= sin(x) times x or sin(x)/x if x is not 0? In the first case, the limit is 0, in the second, 1.

It is important that you understand that in $\lim_{x\to a} f(x)$, the value of f(a) is completely irrelevant! The limit is the value of the function when x is arbitrarily close to a but NOT equal to a. The precise statement of the definition of "$\lim_{x\to a} f(x)$" is:
"$\lim_{x\to a} f(x)= L$" if and only if, for any $\epsilon> 0$, there exist $\delta> 0$ such that if $0< |x- a|< \delta$, then $|f(x)- L|< \epsilon$."

Look specifically at that "0< |x- a|". We are NOT saying anything about what happens when |x- a|= 0, that is, when x= a.

If 0< x< 0.0001, then $10000< 1/x< \infty$ and there are many intervals of length $2\pi$ in which sin(1/x) takes on all values from -1 to 1. The same thing happens for $0< x< \delta$ for any positive $\delta$: $1/\delta< x< \infty$ so sin(1/x) goes (infinitely many times) from -1 to 1.

5. Aug 28, 2013

### brmath

types of functions with no limits

Looking at a graph can give you a very good idea of why a limit might not exist, but in itself is never a proof. Suppose you are looking at lim f(x) as x→ a (a is a finite number). One way to prove there is no limit is to show that the right or left limit is ∞ ( or -∞). But you have to show it -- you can't just say "the graph looks that way". The graph may be misleading.

The case of sin(1/x) as x → 0 is one example of a class of functions that oscillate between discrete values in the vicinity of a (in this case a is 0). You would show there is no limit in these cases by showing that the limit of one subsequence approaching a is different from the limit of another subsequence approaching a. In the case of sin 1/x the first sequence can be the points where sin1/ x = 1; and the second the points where sin 1/x = 0. Both subsequences converge to 0; both have finite limits; and the limits are not the same.

Another example of the above is a function with infinitely many point discontinuities as x → a. For example, you could have f(1/n) = 1 (n an integer) and f(x) = x otherwise. This function will not have a limit at 0, although it will everywhere else.

So you see there are several cases where a limit might not exist. Further f(x) might be a very complicated function that is hard to graph.

If you are dealing with limits because you need to get some math or physics done (as opposed to needing to pass a class), it is worth developing your skills in detecting and proving limits.

6. Aug 28, 2013

### CuriousBanker

So if I were to say "prove that the there is no limit for 1/x as x->0, how would you prove it?

Also, if you were to see a function like 1/x as x->0, how would you "know" that the function tends towards infinity, and not that it doesnt just stop at some large number, like 22109123902102939230293923092913? I mean intuitively is obviosu but how do you prove it?

7. Aug 28, 2013

### brmath

Well, I would say, consider the sequence 1/n. If f(x) = 1/x then f(1/n) = n. Since you can pick n as large as you like you can satisfy the strict mathematical definition of the limit being infinity. That says, basically that if you choose any fixed number N, you can always find an x such that f(x) >N. The essential point is that the N has to be picked first. So you pick the N, I'll pick x = 1/(N+1), and f(x) > N.

Now you do have to worry about x's that are between 1/n and 1/(n+1). Theoretically it could be that there is a sequence of a's going to 0 where f(a) = 1 for every a, or something of that kind. If it could happen, then we really do not have a limit. But in the case of 1/x that cannot happen, because 1/x is continuous. We cannot even jiggle 1/x into having peaks and valleys on its way to infinity, because 1/x is monotonic.