极限函数$f(x)$和数列$ f(1),(2),\ ldots $

An object, say a ball, is moving through space. A friend shows you pictures of this ball, say, every second. There's some point that the ball gets arbitrarily close to as you look at later and later pictures. Does it follow the ball is actually doing the same thing?

Of course not! There is no telling what the ball could be doing in between those times. It turns out that this actually highlights a key idea from calculus. Given a function $f$, the sequence $a_1=f(1)$, $a_2=f(2)$, and so on may converge to something, just like the pictures of the ball. However, $f$ itself may do whatever it wants in between. Something like $|.5-\{x\}|$, where $\{ x \}$ is the fractional part of $x$, serves as an example. For integer $x$ this is always at $.5$ even though it actually oscillates between $.5$ and $0$ as $x$ gets large.

There are countless (in fact, uncountably many) other examples of such functions $f$. Just take some sequence of points $f(1)$,.. $f(n)$ that converge to a limit $L$ and draw a function through those points which does not approach $L$.

The converse, however, is true. If the limit of a function $f(x)$ as $x$ gets large is $L$, then the limit of the sequence $f(1),...f(n)$ is also $L$. This is because for any $e>0$ we can find an $N$ such that $n>N$ makes $L-e<f(n)<L+e$.

If we have a function $f(x)$, then

$\bullet$ $\lim_{x\rightarrow\infty}f(x)$ refers to the limit of the function $f(x)$ as $x\rightarrow\infty$.

$\bullet$ $\lim_{n\rightarrow\infty}f(n)$ refers to the limit of the sequence $f(1), f(2), \ldots$. (The variable $n$ usually only refers to integers.)

Here's the warning: These are not always the same!

One example of such a function is $f(x) = \sin (\pi x)$. For any integer $n$, we have $f(n) = 0$, so $\lim_{n\rightarrow\infty}f(n)$ is 0. But $f(x)$ itself doesn't approach any particular number; it continues to oscillate between -1 and 1 forever.

An even easier way--though it might seem like cheating--is to use a conditional definition:

$f(x) = \left\{ \begin{array}{lr} 0 &\text{, $x$ is an integer}\\ 1 &\text{, otherwise} \end{array} \right.$

For this function also, $\lim_{n\rightarrow\infty}f(n)$ is 0. But $f(x)$ itself doesn't approach any particular number. We can't say $f(x)$ approaches 0, because there are arbitrarily large numbers $x$ with $f(x) = 1$, and we can't say $f(x)$ approaches 1, because there are arbitrarily large numbers $x$ with $f(x) = 0$.

So be careful; $\lim_{x\rightarrow\infty}f(x)$ is not necessarily the same as $\lim_{n\rightarrow\infty}$. (But if they both exist, then they must be the same. Try proving this for a challenge.)

You may have noticed that the definitions of limits for functions at infinity and for sequences are similar. To start off, they both use the word "limit". Is there a connection?

Take some function like $f(x)=\frac{x^2+2}{x^2+1}$

Now we can look at the sequence given by $a_i=f(i)$ . We then get $\\$ $\lim \limits_{n \to \infty} a_ i=\lim \limits_{x \to \infty}f(x)$.

Graphically, we are just selecting integer $x$:

Sometimes, it is easier to think about the limit of a function in terms of the limit of a sequence.

But be careful: the sequence $a_n=f(n)$ could have a limit even if the function $f(x)$ at infinity does not. An example is $f(x)=\cos(2\pi x)$. At integers, $\cos(2\pi n)=1$, so $(a_n)=(1,1,\ldots)$. $\\$ $\text{But } \lim \limits_{x \to \infty} \cos(2\pi x)$ does not exist.