# When do limits at infinity not exist?

Here is a limit at infinity$$\lim_{x \to \infty}f(x)$$

A limit fails to exist for one of the four reasons:

1. The one-sided limits are not equal
2. The function doesn't approach a finite value
3. The function oscillates
4. The $x$ value is approaching the endpoint of a closed interval

Here are my questions:

1. How does this work for limits at infinity? I am under the impression that we can only approach infinity from one side, therefore causing all limit at infinity to fail this test.

2. So the only case where this rule applies is if the limit evaluates to either $+\infty$ or $-\infty$?

3. No questions.

4. Is this case essentially a subset of case 1?

• This is a "limit at infinity", not an "infinite limit". An example of an infinite limit is $\lim\limits_{x\,\downarrow\,0} \dfrac 1 x. \qquad$ – Michael Hardy Sep 17 '16 at 19:03
• Right, limits to $\infty$ are "one-sided" only. So $2$ and $3$ hold. (No idea what you mean with $4$.) – Yves Daoust Sep 17 '16 at 19:35
• "I am under the impression that we can only approach infinity from one side, therefore causing all infinite limits to fail this test." No. That would make all infinite limits pass that test. (but not nescessarily any of the others.) – fleablood Sep 17 '16 at 19:49

I'll try to give some example.

Take the function

$$f(x) = \ln(x)$$

When you're going to compute the limit for $x\to \infty$, you see it doesn't exist. You need to compute both the limits to see it clearly.

$$\lim_{x\to+\infty}\ln(x) = +\infty$$

$$\lim_{x\to-\infty}\ln(x) = \text{doesn't exist}$$

in $\mathbb{R}$ the logarithm is indeed defined for $x > 0$. The value $x = 0$ itself is not well defined, since the only possible limit is $0^+$.

In this way, the rules for the infinities are pretty much the same of those for generic numbers which represents vertical asymptote of a function.

The logarithm example might be the case in which you are approaching to a forbidden zone, namely the zone at the left of zero in which the log doesn't exist.

Another example:

$$g(x) = e^{-x}$$

In this case you have $0$ for $x\to +\infty$ and $+\infty$ for $x\to -\infty$ hence the limit to infinity is not defined either.

In this case you can approach to both sides, because the exponential function is well defined on all the real axis, but as you can see the limits are different.

So, in few words, you have always to check for both the infinities to see if a function is defined.

The next example is for

$$h(x) = \frac{1}{x^2}$$

In this case both $x\to +\infty$ and $x\to -\infty$ give you the same limit, which is zero. That function then possess a limit to infinity.

Those four reasons are somewhat peculiar sounding. The fourth one doesn't even seem correct to me. Take $f:[0,1] \to \mathbb{R}$, $x \mapsto x$. Taking the limit as $x \to 0$ is valid even though its a closed interval.

For finite limits, the limit as $x \to c$ "fails to exist" if for every real number $L$ the following holds: for some $\epsilon >0$ there is no neighbourhood $(c-\delta, c+\delta)$ such that if $x \in (c-\delta, c+\delta) \ \cap \text{dom} \ f$, $|f(x) - L| < \epsilon$. This is precisely the negation of the limit definition.

Now, for limits where $x \to \infty$, again, you can just negate the definition. Here, the idea is, roughly speaking, that you can't make $f(x)$ arbitrarily close to a particular number by making $x$ sufficiently large. Formally, for every $L$, we have the following: for some $\epsilon$ there is no $M$ such that if $x>M$, $|f(x) - L| < \epsilon$.

Here, we assume the value of the limit itself is finite. Infinite limits, where $|f(x)| \to \infty$ as $x \to c$, can be dealt with in a similar manner. You just need precise definitions.