Differential and Integral Calculus 2021

The function \(f(x)=x \sin(1/x)\), \(x>0\) does have the limit \(0\) at \(0\).

The function \(g(x)= \sin(1/x)\), \(x>0\) does not have a limit at \(0\).

An important property of limits is that they are always unique. That is, if \(\lim_{x\to x_0} f(x)=a\) and \(\lim_{x\to x_0} f(x)=b\), then \(a=b\). Although a function may have only one limit at a given point, it is sometimes useful to study the behavior of the function when \(x_k\) approaches the point \(x_0\) from the left or the right side. These limits are called the left and the right limit of the function \(f\) at \(x_0\), respectively.

Definition 2: One-sided limits

Suppose \(S\) is a set in \(\mathbb{R}\) and \(f\) is a function defined on the set \(S\setminus\{x_0\}\). Then we say that \(f\) has a left limit \(y_{0}\) at \(x_{0}\), and write \[\lim_{x \to x_{0}-}f(x)=y_{0},\] if, \(f(x_{k})\to y_{0}\) as \(k\to \infty\) for every sequence \((x_{k})\) in the set \(S\cap ]-\infty,x_0[ =\{ x\in S : x < x_0 \}\), such that \(x_{k}\to x_{0}\) as \(k\to \infty\).

Similarly, we say that \(f\) has a right limit \(y_{0}\) at \(x_{0}\), and write \[\lim_{x \to x_{0}+}f(x)=y_{0},\] if, \(f(x_{k})\to y_{0}\) as \(k\to \infty\) for every sequence \((x_{k})\) in the set \(S\cap ]x_0,\infty[ =\{ x\in S : x_0 < x \}\), such that \(x_{k}\to x_{0}\) as \(k\to \infty\).

The sign function \[\mathrm{sgn}(x)= \frac{x}{|x|}\] is defined on \(S= \mathbb{R}\setminus 0\). Its left and right limits at \(0\) are \[\lim_{x\to 0-} \mathrm{sgn}(x)= -1,\qquad \lim_{x\to 0+} \mathrm{sgn}(x)= 1.\] However, the function \(\mathrm{sgn}(x)\) does not have a limit at \(0\).

Example 7.

Function \(f: \mathbb{R}\setminus 0 \to \mathbb{R}\) \[f(x) = \frac{1}{x}\] does not have one-sided limits at 0.

The following limit rules are immediately obtained from the definition and basic algebra of real numbers.

Example 8.

Finding limits by calculating \(f(x_0)\):

a) \[\lim_{x\to 2}(5x-3)=10-3=7.\]

b) \[\lim_{x\to -2}\frac{3x+2}{x+5} = \frac{-6+2}{-2+5}=-\frac{4}{3}.\]

c) \[\lim_{x\to 2} \frac{x^2-4}{x-2} = \lim_{x\to 2} \frac{(x+2)(x-2)}{x-2} = \lim_{x\to 2}(x+2) = 4.\]

For continuity of a function \(f\) at a given point \(x_0\), it is required that:

1. \(f(x_0)\) is defined,

2. \(\lim_{x \to x_0} f(x)\) exists (and is finite),

3. \(\lim_{x \to x_0} f(x) = f(x_0)\).

In other words:

Let \(c\in \mathbb{R}\). Functions \(f,g,h\) defined by \(f(x)=c\), \(g(x)=x\), \(h(x)=|x|\) are continuous at every point \(x\in \mathbb{R}\).

Why? If \(x_{k}\to x_{0}\), then \(f(x_{k})=c\) and \(\lim_{k\to \infty}f(x_k)= c=f(x_{0})\). For \(g\), we have \(g(x_{k})=x_{k}\) and hence, \(\lim_{k\to\infty} g(x_k)=x_{0}=g(x_{0})\). Similarly, \(h(x_{k})=|x_{k}|\) and \(\lim_{k\to\infty}h(x_k)= |x_{0}|=h(x_{0})\).

Let \(x_{0}\in \mathbb{R}\). We define a function \(f\colon\mathbb{R}\to \mathbb{R}\) by \[f(x)= \left\{\begin{array}{rl}2 & \text{ for }x \lt x_{0}, \\ 3 & \text{ for }x\geq x_{0}.\end{array}\right.\] Then \[\lim_{x \to x_{0}^{-}}f(x)=2,\text{ and } \lim_{x \to x_{0}^{+}}f(x)=3.\] Therefore \(f\) is not continuous at the point \(x_{0}\).

Some basic properties of continuous functions of one real variable are given next. From the limit rules (Theorem 2) we obtain:

A composition of continuous functions is continuous if it is defined:

Proof.

Note. If \(f\) is continuous, then \(|f|\) is continuous.

Why?

Write \(g(x):=|x|\). Then \((g\circ f)(x)=|f(x)|\).

Note. If \(f\) and \(g\) are continuous, then \(\max (f,g)\) and \(\min (f,g)\) are continuous. (Here \(\max (f,g)(x):=\max \{f(x),g(x)\}\).)

Why?

Write \[\begin{cases}(a+b)+|a-b|=2\max(a,b), \\ (a+b)-|a-b|=2\min(a,b). \end{cases} \]

The so-called \((\varepsilon,\delta)\)-definition for continuity is given next. The basic idea behind this test is that, for a function \(f\) continuous at \(x_0\), the values of \(f(x)\) should get closer to \(f(x_0)\) as \(x\) gets closer to \(x_0\).

This is the standard definition of continuity in mathematics, because it also works for more general classes of functions than ones on this course, but it not used in high-school mathematics. This important definition will be studied in-depth in Analysis 1 / Mathematics 1.

\((\varepsilon,\delta)\)-test:

Proof.

From Theorem 3 we already know that the function \(f: \mathbb{R} \to \mathbb{R}\) defined by \(f(x) = 4x\) is continuous. We can also use the \((\varepsilon,\delta)\)-definition to prove this.

Let \(x_{0}\in \mathbb{R}\). We define a function \(f\colon\mathbb{R}\to \mathbb{R}\) by \[f(x)= \left\{\begin{array}{rl}2 & \text{ for }x \lt x_{0}, \\ 3 & \text{ for }x \geq x_{0}.\end{array}\right.\] In Example 2 we saw that this function is not continuous at the point \(x_0\). To prove this using the \((\varepsilon,\delta)\)-test, we need to find some \(\varepsilon > 0\) and some \(x_\delta \in \mathbb{R}\) such that for all \(\delta > 0\), \(|x_\delta - x_0| < \delta\), but \(|f(x_\delta) - f(x_0)| > \varepsilon\).