Limits and Infinity

One of the mysteries of Mathematics seems to be the concept of "infinity", usually denoted by the symbol $\infty$. So what is $\infty$? It is simply a symbol that represents large numbers. Indeed, numbers are of three kinds: large, normal size, and small. The normal size numbers are the ones that we have a clear feeling for. For example, what does a trillion mean? That is a very large number. Also numbers involved in macro-physics are very large numbers. Small numbers are usually used in micro-physics. Numbers like 10-75 are very small. Being positive or negative has special meaning depending on the problem at hand. The common mistake is to say that $-\infty$ is smaller than 0. While this may be true according to the natural order on the real line in term of sizes, $-\infty$ is big, very big!

So when do we have to deal with $\infty$ and $-\infty$? Easy: whenever you take the inverse of small numbers, you generate large numbers and vice-versa. Mathematically we can write this as:

\begin{displaymath}\frac{1}{0} = \pm \infty\;\;\mbox{and}\;\; \frac{1}{\infty} = 0\;.\end{displaymath}

Note that the inverse of a small number is a large number. So size-wise there is no problem. But we have to be careful about the positive or negative sign. We have to make sure we know whether a small number is positive or negative. 0+ represents small positive numbers while 0- represents small negative numbers. (Similarly, we will use e.g. 3+ to denote numbers slightly bigger than 3, and 3- to denote numbers slightly smaller than 3.) In other words, being more precise we have

\begin{displaymath}\frac{1}{0+} = +\infty\;\;\mbox{and}\;\; \frac{1}{0-} = -\infty\;.\end{displaymath}

Remark. Do not treat $\pm\infty$ as ordinary numbers. These symbols do not obey the usual rules of arithmetic, for instance, $\infty +1=\infty$, $\infty -1=\infty$, $2 \cdot\infty=\infty$, etc.

Example. Consider the function

\begin{displaymath}f(x) = \frac{1}{x-3}\;\cdot\end{displaymath}

When $x \rightarrow 3$, then $x-3 \rightarrow 0$. So

\begin{displaymath}\lim_{x \rightarrow 3-} f(x) = \frac{1}{0-} = -\infty\;\;\mbo...
\lim_{x \rightarrow 3+} f(x) = \frac{1}{0+} = +\infty\;.\end{displaymath}

Note that when x gets closer to 3, then the points on the graph get closer to the (dashed) vertical line x=3. Such a line is called a vertical asymptote. For a given function f(x), there are four cases, in which vertical asymptotes can present themselves:

$\displaystyle \lim_{x \rightarrow a-} f(x) =
-\infty$;     $\displaystyle \lim_{x \rightarrow a+} f(x) =
$\displaystyle \lim_{x \rightarrow a-} f(x) =
-\infty$;     $\displaystyle \lim_{x \rightarrow a+} f(x) =
$\displaystyle \lim_{x \rightarrow a-} f(x) =
+\infty$;     $\displaystyle \lim_{x \rightarrow a+} f(x) =
$\displaystyle \lim_{x \rightarrow a-} f(x) =
+\infty$;     $\displaystyle \lim_{x \rightarrow a+} f(x) =

Next we investigate the behavior of functions when $x \rightarrow
\pm\infty$. We have seen that $\displaystyle \frac{1}{\pm \infty}
= 0$. So for example, we have

\begin{displaymath}\lim_{x \rightarrow -\infty} \frac{1}{x} = 0\;\;\mbox{and}\;\;
\lim_{x \rightarrow +\infty} \frac{1}{x} = 0\;.\end{displaymath}

In the next example, we show how this result is very useful.

Example. Consider the function

\begin{displaymath}f(x) = \frac{2x+1}{x-1}\;\cdot\end{displaymath}

We have

\begin{displaymath}\frac{2x+1}{x-1} = \frac{2+\displaystyle \frac{1}{x}}{1-\displaystyle \frac{1}{x}}\end{displaymath}

which implies

\begin{displaymath}\lim_{x \rightarrow \pm \infty} f(x)= \frac{2+0}{1-0} = 2\;.\end{displaymath}

Note that when x gets closer to $\pm\infty$ (x gets large), then the points on the graph get closer to the horizontal line y=2. Such a line is called a horizontal asymptote.

In particular, we have

\begin{displaymath}\lim_{x \rightarrow -\infty} \frac{a}{x^r} = 0\;\;\mbox{and}\;\;
\lim_{x \rightarrow +\infty} \frac{a}{x^r} = 0\end{displaymath}

for any number a, and any positive number r, provided xr is defined. We also have

\begin{displaymath}\lim_{x \rightarrow \infty} x^r = \infty\;.\end{displaymath}

For $-\infty$, we have to be careful about the definition of the power of negative numbers. In particular, we have

\begin{displaymath}\lim_{x \rightarrow -\infty} x^n = (-1)^n \infty \end{displaymath}

for any natural number n.

Example. Consider the function

\begin{displaymath}f(x) = \frac{2x^4 -3x^2 + 5}{3x^4 + 2x +5}\;\cdot\end{displaymath}

We have

\begin{displaymath}\frac{2x^4 -3x^2 + 5}{3x^4 + 2x +5} = \frac{\displaystyle 2 -...
...5}{x^4}}{\displaystyle 3 + \frac{2}{x^3} +\frac{5}{x^4}}\;\cdot\end{displaymath}

So we have

\begin{displaymath}\lim_{x \rightarrow \pm \infty} f(x) = \frac{2 -0 + 0}{3 + 0 +0} = \frac{2}{3}\;\cdot\end{displaymath}

Example. Consider the function

\begin{displaymath}f(x) = \frac{\sqrt{4x^2 +2}}{3x+1}\;\cdot\end{displaymath}

We have

\begin{displaymath}\sqrt{4x^2 + 2} = \sqrt{x^2\left(4 + \frac{2}{x^2}\right)} = \vert x\vert \sqrt{4 + \frac{2}{x^2}}\end{displaymath}

and then

\begin{displaymath}\frac{\sqrt{4x^2 +2}}{3x+1} = \frac{\vert x\vert}{x} \frac{\d...
...ystyle \sqrt{4 + \frac{2}{x^2}}}{\displaystyle 3 + \frac{1}{x}}\end{displaymath}

When x goes to $+\infty$, then x > 0, which implies that |x| = x. Hence

\begin{displaymath}\lim_{x \rightarrow +\infty} f(x) = \frac{\sqrt{4 + 0}}{3 + 0} = \frac{2}{3}\;\cdot\end{displaymath}

When x goes to $-\infty$, then x < 0, which implies that |x| = -x. Hence

\begin{displaymath}\lim_{x \rightarrow -\infty} f(x) = -\frac{\sqrt{4 + 0}}{3 + 0} = -\frac{2}{3}\;\cdot\end{displaymath}

Remark. Be careful! A common mistake is to assume that $\sqrt{x^2} = x$. This is true if $x \geq 0$ and false if x < 0.

Exercise 1. Find

\begin{displaymath}\lim_{x \rightarrow \infty} \sqrt[3]{\frac{x^2+3}{27x^2 -1}}\;\cdot\end{displaymath}


Exercise 2. Find

\begin{displaymath}\lim_{x \rightarrow -\infty} \frac{x-2}{\sqrt{x^2 +1}}\;\cdot\end{displaymath}


Exercise 3. Find

\begin{displaymath}\lim_{x \rightarrow \infty} \frac{x^2+1}{2x-3}\;\cdot\end{displaymath}


Exercise 4. Find

\begin{displaymath}\lim_{x \rightarrow \infty} 2x+1 - \sqrt{4x^2 + 5}\;.\end{displaymath}


Exercise 5. Find the vertical and horizontal asymptotes for the graph of

\begin{displaymath}f(x) = \frac{\sqrt{x^2+1}}{x-2}\;\cdot\end{displaymath}


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