Infinity

Instead of investigating what happens with a function as its argument approaches a real number, one can ask how a function behaves as its argument increases and remains larger than any real number, or as it decreases and remains less than any real number.

We will say that a quantity ``approaches positive infinity'' if it eventually exceeds and remains larger than any given number. We will say that a quantity ``approaches negative infinity'' if it eventually falls below and remains less than any given number.

We may then ask: ``If $f:(-\infty,\infty)\rightarrow(-\infty,\infty)$, what happens to f(x) as x approaches positive infinity?'' If f(x) approaches a real number L, we will write

$$\lim_{x\rightarrow +\infty}f(x) = L.$$

For example, we would say if $f:(-\infty,\infty)\rightarrow(-\infty,\infty)$and f(x) = x²/(1+x²) then since

$$f(x) = \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2}$$

we would say that

$$\lim_{x\rightarrow\infty} f(x) = 1 - 0 = 1$$

This concept is probably familiar to you from your study of horizontal asymptotes.

Notice that it is important that we have both that the quantity exceed any real number and that it eventually remain larger than any given real number before we conclude that it approaches infinity. For example, if $f:(-\infty,\infty)\rightarrow(-\infty,\infty)$ and $f(x) = x\sin(x)$,we would not be justified in saying that f(x) approaches positive infinity as x approaches infinity, since for every x of the form $k\pi$ where k is an integer, f(x) = 0. The values of f continue to oscillate between ever larger positive values and ever more negative values, as x approaches positive infinity.

Similarly, if $f:(-\infty,0)\cup(0,\infty)\rightarrow(-\infty,\infty)$ with f(x) = 1/x³, we would not be justified in saying that f(x) approaches positive infinity as x approaches since for negative values of x we see that f(x) does not even exceed , let alone all real numbers.

By contrast, if $g:(-\infty,\infty)\rightarrow(0,\infty)$ with g(x) = 1/x², we would be justified in saying that g(x) approaches positive infinity as x approaches . In such cases we will write

$$\lim_{x\rightarrow 0} g(x) = +\infty$$

Notice that this is completely different sort of assertion then

$$\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1$$

insofar as 1 is a number and $+\infty$ is not. The statement that

$$\lim_{x\rightarrow 0}\frac{1}{x^2} = +\infty$$

is simply a refinement of the statement that there is no number which 1/x² approaches as x approaches . It provides an explanation as to why this happens: 1/x² exceeds and remains above any given real number as x approaches . How then could 1/x² approach 10, say, if we know that it will exceed and remain above 100?

Again, from your college algebra classes you are familiar with this behaviour from your study of asymptotes for graphs of equations such as

$$\begin{array} {rcl} y & = & {\displaystyle \frac{x^2 +1}{x(x+1)}}\\ y & = & {\displaystyle \frac{x +11}{x(x+1)}}\end{array}$$

There you learned to identify asymptotes, both vertical and horizontal, by a combination of factoring and long division.

Here then are two examples where factoring will get us no where:

Example 1:

Suppose $y = \sqrt{x^2 + 1} - x$. Determine if the graph of this equation has any horizontal asymptotes. This is tough since we seem to be subtracting one large number from another. We do see that since $\sqrt{x^2 + 1} \gt \sqrt{x^2} = \vert x\vert$ that y > 0 regardless of x. We can do a little better:

$$y = \sqrt{x^2 + 1}-x\times\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x} =\frac{1}{\sqrt{x^2+1}+x}$$

Now we see that as x approaches positive infinity that y approaches . Here is picture of part of the graph of $y = \sqrt{x^2+x} - x$.

What happens as x approaches negative infinity?

Example 2:

Suppose $y = \sqrt{x^2+x} - x$. Determine if the graph of this equation has any horizontal asymptotes.

This is even harder than the last one, as now we see that as x increases through positive values, the difference between $\sqrt{x^2+x}$ and x seems to grow. One might guess here that y approaches positive infinity as x does. The algebra reveals otherwise:

$$y = \sqrt{x^2 + x}-x\times\frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x} =\frac{x}{\sqrt{x^2+1}+x} = \frac{1}{\sqrt{1+(1/x)} + 1}$$

Now we see that as x approaches positive infinity that y approaches 1/2, and, in fact y increases to 1/2. Here is picture of part of the graph of $y = \sqrt{x^2+x} - x$.