# Why are there limits which do not exist?

The limit of a real function can be of one of three types.

A real limit would be a limit where the value is a real number: $$\lim_{x\to0}x^2=0$$ An infinite limit would be a limit where the value is equal to $\pm\infty$: $$\lim_{x\to-\infty}{1\over e^x}=\infty$$ And an undefined/non-existing limit would be one where it is not possible to assign a value of any kind to the limit: $$\lim_{x\to\infty}\sin x$$

My question is, what other undefined limits are out there? I understand that $\lim_{x\to\infty}\sin x$ is undefined because $\sin x$ is a periodic function whose value is in $[-1,1]$, and we can't find which, if any, single value in that interval it will achieve as $x$ tends to infinity. I also know I could get a similar undefined limit using any other trigonometric function.

Are there any other limits composed of elementary functions which are undefined? Are such undefined limits something exclusive to periodic functions or can they be constructed using other kinds of functions?

• The function $\sin(1/x)$ is a good one to be familiar with, by the way. – Akiva Weinberger Nov 18 '15 at 1:49
• It is much better to consider infinite limits as a case of non-existence of limit. The symbol $\infty$ does not make much sense as a number and saying that "value is equal to $\pm\infty$" leads to more confusion. However there are many authors who follow the convention you have mentioned in your post. – Paramanand Singh Nov 18 '15 at 4:12

## 5 Answers

The limits that do not exist can be seen as the set of accumulation points.

In this sense, $\displaystyle\lim_{x\to\infty}\sin x = [-1,1]$ because for each $L\in [-1,1]$ there is a sequence $x_n \to \infty$ such that $\sin(x_n) \to L$.

So, more precisely, the set of accumulation points is the set of all possible limits that do exist.

How about: $$\lim_{x\to0}\frac1{1+3^{1/x}}$$ A good online graphing calculator can be found here.

One of the very simple ways a limit fails to exist is when the left hand and right hand limits exist but are not equal. Thus the following example $$\lim_{x \to 0}\frac{\sqrt{x^{2}}}{x}$$ is a very good one. The left hand limit is $-1$ and the right hand limit is $-1$.

It is easy to design cases when even one sided limits also don't exist. Thus $\sin (1/x)$ oscillates finitely when $x \to 0$ and $(1/x)\sin (1/x)$ oscillates infinitely when $x \to 0$.

Periodicity is not necessary to construct such example as you might seem to think. But if you remove elementary periodic functions from the list then we need to artificially craft examples for non-existence of limit (say by using $|x|$ or $[x]$ or something similar).

How about $$f(x) = \chi_\mathbb{P}(x)$$ thus $1$ if $n$ is integer and prime and otherwise $0$. I believe it is non periodic and has no limit.

• But the set of accumulation points is $\{0,1\}$, which describes clearly what this sequence does. – lhf Nov 18 '15 at 1:38

What you'll usually see is limits as $x$ approaches some constant (typically $0$) that don't exist at vertical asymptotes (typically seen in rational functions). Example:

$$\displaystyle\lim_{x\to\ -2}\ \frac{1}{x+2}$$

Be careful that it's actually a vertical asymptote, and not a removable discontinuity, like the following:

$$\displaystyle\lim_{x\to\ 2} \frac{(x+2)^2}{x+2}$$

As far as undefined limits at infinity, these are often seen with trig functions because they are periodic. Over any interval $\pi$, $\tan(x)$ is defined at all $y$ values, thus its range is $(-\infty,\infty)$. Since tangent is periodic, then as $x$ approaches $\infty$, $\tan(x)$ could be defined for any $y$ value, so its limit as $x$ approaches $\infty$ is indeterminate.

Let me know if you have any questions, and I hope this helps clarify.