The limit of a function

First example and intuitive definition

  • Studying limits means that we investigate the values of \(f(x)\) as \(x\) approaches a given number \(a\)
    • Let's do this in case \(f(x)=x^2-x+2\) and \(a=2\).
    • Click here for a table of values and a graph
    • You can see that the closer \(x\) gets to \(2\), the closer \(f(x)\) gets to \(4\).
    • It appears that we could make \(f(x)\) get arbitrarily close to \(4\) by making \(x\) get arbitrarily close to \(2\).
    • We express this by saying that the limit of the function \(f(x)=x^2-x+2\) as \(x\) approaches 2 is 4.
    • The notation for this is \(\lim_{x\to a}f(x)=4\).
  • Intuitive definition of a limit.
    • Suppose that a function \(f(x)\) is defined near a number \(a\). This means that there exists an open interval \((\alpha,\beta)\) around \(a\) such that \(f(x)\) is defined whenever \(\alpha<x<\beta\), except possibly \(a\).
    • Then we write \(\lim_{x\to a}f(x)=L\) and say that the limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\), if we can make sure that \(f(x)\) is arbitrarily close to \(L\) by making \(x\) close enough to \(a\), but not equal to \(a\).
    • For example, on the table and the graph, you can see that if \(1.999<x<2.001\), then \(3.997<f(x)<4.003\).
    • An alternative notation for \(\lim_{x\to a}f(x)=L\) is \(f(x)\to L\text{ as }x\to a\).

Example: \(\lim_{x\to1}\frac{x-1}{x^2-1}\)

  • Consider the following list of values for \(f(x)=\frac{x-1}{x^2-1}\).
x f(x) x f(x)
0.9 0.526315789474 1.1 0.47619047619
0.999 0.500250125063 1.001 0.499750124938
0.99999 0.500002500013 1.00001 0.499997500012
0.9999999 0.50000002498 1.0000001 0.49999997502
  • Our guess is that \(\lim_{x\to1}\frac{x-1}{x^2-1}=\frac12\), and that is indeed the case.
    • Click here for a graph.
    • Note that by definition the limit \(\lim_{x\to a}(f(x))\) not depend on whether \(f(x)\) is defined at \(x=a\), or what \(f(x=a)\) is. For example, taking \[ g(x)=\begin{cases} \frac{x-1}{x^2-1} & x\ne1 \\ 2 & x=1, \end{cases} \] we still get \(\lim_{x\to1}g(x)=\frac12\).

The perils of subtraction

  • Consider the following list of values for \(f(t)=\frac{\sqrt{t^2+9}-3}{t^2}\).
t f(t) t f(t)
-0.1 0.1666203960726864 0.1 0.1666203960726864
-0.001 0.1666666620547819 0.001 0.1666666620547819
-0.0000001 0.1665334536937735 0.0000001 0.1665334536937735
-0.00000000001 0 0.00000000001 0
  • Up until \(x=\pm0.000001=\pm10^{-5}\), it looks like \(\lim_{t\to0}\frac{\sqrt{t^2+9}-3}{t^2}=1.6666\ldots=\frac16\), but then the numbers drop to 0.
    • Click here for the corresponding plots, where you can see a similar behaviour.
    • As we will be able to prove, it is indeed the case that \(\frac{\sqrt{t^2+9}-3}{t^2}\to\frac16\) as \(x\to\frac16\).
    • The zeroes you get for \(x\) very close to 0 come from the limitations of numerical calculations.
    • Every calculator or computer program performs numerical calculations with numbers rounded up to a fixed number.
    • For example, in case of SAGE, the program I'm using, this number format is that of double-precision floating points: numbers are stored in 8 bytes.
    • The point is that because of this rounding procedure, when the nominator \(\sqrt{t^2+9}-3\) gets really close to zero, it is treated as being 0, before getting divided by \(t^2\).
    • See more in the pdf Lies my calculator and computer told me on the textbook's webpage.

\(\lim_{x\to0}\frac{\sin(x)}{x}\)

  • Consider the following list of values for \(f(x)=\frac{\sin x}{x}\).
x f(x) x f(x)
-0.1 0.9983341664682815 0.1 0.9983341664682815
-0.001 0.9999998333333416 0.001 0.9999998333333416
-0.00001 0.9999999999833333 0.00001 0.9999999999833333
-0.0000001 0.9999999999999983 0.0000001 0.9999999999999983
  • Our guess is that \(\lim_{x\to0}\frac{\sin(x)}{x}=1\), which is indeed the case: we'll see how to prove this in Chapter 3.

\(\lim_{x\to0}\sin\frac{\pi}{x}\)

  • Consider the following list of values for \(f(x)=\sin\frac{\pi}{x}\).
x f(x) x f(x)
-0.1 0.0000000000000001 0.1 -0.0000000000000001
-0.001 -0.0000000000000196 0.001 0.0000000000000196
-0.00001 0.0000000000012188 0.00001 -0.0000000000012188
  • We're getting numbers really close to 0, but they don't seem to get closer to 0 when \(x\) does.
    • That itself is a result of numerical approximation: we can calculate these values precisely \[ f(\pm0.1)=\sin(\pm10\pi)=0,\,f(\pm0.001)=\sin(\pm1000\pi)=0,\,f(\pm0.00001)=\sin(100000\pi)=0. \]
    • This could lead you to think that the limit is zero.
    • But then, you can try another set of numbers getting arbitrarily close 0 zero, which give a different result \[ f(\frac{1}{10.5})=\sin(10.5\pi)=1,\,f(\frac{1}{1000.5})=\sin(1000.5\pi)=1,\,f(\frac{1}{100000.5})=\sin(100000.5\pi)=1. \]
    • That is, it is not true that \(\lim_{x\to}\sin\frac{\pi}{x}=0\), because we can find numbers \(x\) arbirarily close to 0, for which \(f(x)=1\), and vice versa.
    • This means that \(\lim_{x\to0}\sin\frac{\pi}{x}\) does not exist.
    • Click here for a graph.

One-sided limits

  • What could be said about \(f(x)=\sqrt x\) as \(x\) approaches 0?
    • The issue is that the function \(f(x)\) is defined on \([0,\infty)\), so you can only approach \(0\) from the right.
    • It looks like we can make sure that \(f(x)\) is arbitrarily close to \(0\) if \(x\) is sufficiently close to \(a\) with \(x>a\):
x f(x)
0.1 0.3162277660168379
0.001 0.0316227766016838
0.00001 0.0031622776601684
  • This is an example of a one-sided limit. Click here to see a graph.
    • Definition of right-hand limit. Suppose that \(f(x)\) is defined on an open interval having \(a\) as a left endpoint. We say that the right-hand limit of \(f(x)\) as \(x\) approaches \(a\) (or the *limit of \(f(x)\) as \(x\) approaches \(a\) from the right) is \(L\), and we write \[ \lim_{x\to a^+}f(x)=L, \] if we can make sure that \(f(x)\) is arbitrarily close to \(L\) by taking \(x\) sufficiently close to \(a\) with \(x>a\).
    • Similarly, we can define left-hand limits. The notation for those is \[ \lim_{x\to a^-}f(x)=L. \]

\(\lim_{x\to0}\frac{1}{x^2}\)

  • Consider the following list of values for \(f(x)=\frac{1}{x^2}\).
x f(x) x f(x)
-0.1 100 0.1 100
-0.001 1000000 0.001 1000000
-0.00001 10000000000 0.00001 10000000000
  • You can see that as \(x\) approaches \(0\), the value \(f(x)\) gets arbitrarily large.
    • Click here for a graph
    • We denote this behaviour by \(\lim_{x\to0}\frac{1}{x^2}=\infty\).
    • Intuitive definition of an infinite limit. Let \(f(x)\) be a function defined around the number \(a\), except possibly \(a\) itself. Then we write \[ \lim_{x\to a}f(x)=\infty, \] if we can make sure that the value \(f(x)\) gets arbitrarily large, by taking \(x\) sufficiently close to \(a\).
    • Note that \(\infty\) is not a number, the limit does not exist. But this nonexistence happens in a particular way, which is indicated by the notation \(\lim_{x\to a}f(x)=\infty\).
    • Similarly, if we can make sure that the value \(f(x)\) gets arbitrarily small by taking \(x\) sufficiently close to \(a\), we write \(\lim_{x\to a}f(x)=-\infty\). For a cheap example, consider \(f(x)=-\frac{1}{x^2}\).

\(\lim_{x\to(\pi/2)^-}\tan(x)\) and \(\lim_{x\to(\pi/2)^+}\tan(x)\)

  • Consider the following values for \(f(x)=\tan(x)\).
x f(x) x f(x)
\(\pi/2-0.1\) 9.9666444232592362 \(\pi/2+0.1\) -9.9666444232592379
\(\pi/2-0.001\) 999.999666666637836 \(\pi/2+0.001\) -999.9996666666493184
\(\pi/2-0.00001\) 99999.9999966482282616 \(\pi/2+0.00001\) -99999.9999967626499711
  • It looks like \(\lim_{x\to(\pi/2)^-}\tan(x)=\infty\) and \(\lim_{x\to(\pi/2)^+}\tan(x)=-\infty\).
    • Click here for the graph.
    • You can see that as you approach \(\frac{\pi}{2}\) with \(x\), the tangent line to the graph \(y=\tan(x)\) gets closer and closer to the vertical line \(x=\frac{\pi}{2}\).
    • In this case, we say that the line \(x=\frac{\pi}{2}\) is a vertical asymptote of \(\tan(x)\).
    • Definition. We say that the line \(x=a\) is a vertical asymptote of the curve \(y=f(x)\), if at least one of the following statements is true. \[\begin{gather*} \lim_{x\to a}f(x)=\infty\quad\lim_{x\to a^-}f(x)=\infty\quad\lim_{x\to a^+}f(x)=\infty\\ \lim_{x\to a}f(x)=-\infty\quad\lim_{x\to a^-}f(x)=-\infty\quad\lim_{x\to a^+}f(x)=-\infty. \end{gather*}\]
    • Exercises. 2.2: 4, 6, 8, 16, 20, 32, 52