Numbers are not always simple. If you are told to "count as far as you can," how far can you go? The truth is that you can go on forever. For every number, there is a bigger one. Infinity (denoted \(\infty\)) is of a value greater than all numbers that is not itself a number. It basically refers to "the end of counting," and has many applications that the finite world cannot handle. Some of those applications will be looked at in this article.
A limit refers to the value that is approached for a certain value of an argument for a function. A limit can be finite or infinite; in other words, it either converges or diverges. In many cases looking at graphs helps a lot.
We begin by analyzing a classic example from trigonometry in terms of limits.
Example 1: What is \(\lim_{x \rightarrow \frac{\pi}{2}} \tan x \)?
Solution: Graphing \(\tan x\) will make it clear that there is a vertical asymptote at \(x = \frac{\pi}{2}\). However, on one side of the asymptote the graph approaches negative infinity, and on the other side it approaches positive infinity. We cannot determine what the true limit is, so we say that it does not exist (saying that it diverges is also acceptable).
Here is another example that is very common. It relies on something known as asymptotes, lines with special slopes that the graph of a function mimics. A horizontal asymptote is a horizontal line with slope of \(0\) and an equation \(y = a\), where \(a\) is a finite constant.. Generally a graph will approach horizontal asymptotes only as the argument approaches positive or negative infinity. A vertical asymptote is a vertical line with an undefined slope. A vertical asymptote will always be a line with the equation \(x = b\), where \(b\) is also a finite constant. The function is always undefined at the vertical asymptote.
Example 2: Explain why \(\lim_{n \rightarrow \infty} \frac{1}{2^n} = 0\) using asymptotes.
Solution: Graphing this function will reveal that \(y = 0\) is a horizontal asymptote for the function, and as \(n\) gets bigger, the value of the function approaches \(0\). Therefore \(\lim_{n \rightarrow \infty} \frac{1}{2^n} = 0\).
Here is another example that focuses specifically on vertical asymptotes.
Example 3: What are all vertical asymptotes of the function \(f(x) = \frac{1}{x^3 - x}\)?
Solution: At vertical asymptotes, the function is undefined, so the denominator must be zero. In other words, we must find the roots of the denominator. Thus we solve the equation
$$x^3 - x = 0$$
To do this, factor it in two steps:
$$x(x^2 - 1) = 0 \Rightarrow$$ $$x(x + 1)(x - 1) = 0$$
Thus the roots and the vertical asymptotes are \(x = 0\), \(x = -1\), and \(x = 1\).
Examples 2 and 3 suggest a more formal way of finding horizontal and vertical asymptotes:
To find all horizontal asymptotes, take the limit of the function as it approaches both positive and negative infinity.
To find all vertical asymptotes, find all places where the function is undefined--this will be all roots of the denominator.
The next example uses this process to find both horizontal and vertical asymptotes.
Example 4: What are all horizontal and vertical asymptotes of the function \(f(x) = \frac{2x^2 + 10x - 37}{x^2 + 5x - 24}\)?
Solution: Generally vertical asymptotes are easier to find--they are the roots of the denominator:
$$x^2 + 5x - 24 = 0$$
Factor this equation to solve for \(x\):
$$(x + 8)(x - 3) = 0$$
So the roots are \(x = -8\) and \(x = 3\). These are also the vertical asymptotes.
The horizontal asymptote(s) may appear very tricky to find, but observe the following algebraic manipulations on the function \(f(x)\):
$$f(x) = \frac{2x^2 + 10x - 37}{x^2 - 5x - 24} \Rightarrow$$ $$f(x) = \frac{2x^2 + 10x - 48}{x^2 - 5x - 24} + \frac{11}{x^2 - 5x - 24} \Rightarrow$$ $$f(x) = \frac{2(x^2 + 5x - 24)}{x^2 - 5x - 24} + \frac{11}{x^2 - 5x - 24} \Rightarrow$$ $$f(x) = 2 + \frac{11}{x^2 - 5x + 24}$$
When taking limits of \(x\), the constant term remains the same, and as \(x\) approaches either positive or negative infinity, the denominator of the fraction component grows towards infinity, and thus the fraction approaches zero. Therefore the only horizontal asymptote is \(y = 2\).
In precalculus and calculus classes, a common scenario is to be given a series and to be asked whether it converges or diverges. There are far too many techniques (known as convergence tests) to go over in one article, but a common non-Calculus based one is for geometric series, where the consecutive terms change by a constant factor.
The Geometric Series Test states that for a geometric series with a constant factor of \(r\), the sum converges if \(|r| < 1\) and diverges otherwise.
Here are two examples of the Geometric Series Test in action.
Example 5: Why does the sum \(\frac{1}{3} + \frac{7}{18} + \frac{49}{108} + ...\) diverge?
Solution: Consecutive terms differ by a factor of \(\frac{7}{6} \), and since this is greater than 1, the sum diverges by the Geometric Series test.
Example 6: What is the smallest value of \(n\) such that the series \(3 + \frac{9}{2n} + \frac{27}{4n^2} + \frac{81}{8n^3} + ...\) diverges?
Solution: This is a geometric series with an initial term of \(3\) and a common ratio of \(\frac{3}{2n}\). Since by the geometric series the sum converges if the common ratio is less than \(1\) and greater than \(-1\), the values of \(n\) that will make the series diverge are the solutions to the inequalities
$$\frac{3}{2n} \geq 1, \frac{3}{2n} \leq -1$$
We want the smallest solution, so solve this for \(n\), first by taking the reciprocal of both sides of the inequalities (and reversing its sign):
$$\frac{2n}{3} \leq 1, \frac{2n}{3} \geq -1$$
Now combine the two inequalities into one:
$$-1 \leq \frac{2n}{3} \leq 1$$
Isolate \(n\) by multiplying all three parts of the inequality by \(\frac{3}{2}\):
$$-\frac{3}{2} \leq n \leq \frac{3}{2}$$
The value \(-\frac{3}{2}\) is the smallest number in this interval, so it is the solution we are looking for. Notice that this solution makes the common ratio equivalent to \(-1\), so by the Geometric Series Test it diverges.