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.
• 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.

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.
• 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 $$\lim_{x\to(\pi/2)^+}\tan(x)=-\infty$$.
• 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. $\begin{gather*} \lim_{x\to a}f(x)=\infty\quad\lim_{x\to a^-}f(x)=\infty\quad\lim_{x\to a^+}f(x)=\infty\\ \lim_{x\to a}f(x)=-\infty\quad\lim_{x\to a^-}f(x)=-\infty\quad\lim_{x\to a^+}f(x)=-\infty. \end{gather*}$