Continous functions

Continuous functions Suppose $f(x)$ is a function defined on the interval $(x_{0} - \delta, x_{0} + \delta)$ and satisfies the condition

$\displaystyle \lim_{x \rightarrow x_{0}}f(x) = f(x_{0})$

Then

is continous at $x = x_{0}$ .

$(x_{0} - \delta, x_{0} + \delta)$ The interval $(x_{0} - \delta, x_{0} + \delta)$ is centered at $x_0$ and the distance from $x_0$ is $\delta$ .

NOTE Suppose the domain of $f(x)$ contains the interval $(x_{0} - \delta, x_{0} + \delta)$ . Then $f(x)$ is continuous at $x_0$ except the following two cases.

$\lim_{x \to x_{0}}f(x)$ does not exist
$\lim_{x \to x_{0}}f(x)$ exists but not equal the value $f(x_{0})$

**Figure 1.42:** Discontinuous
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Case 1. $x_{0}$ is called essentail discontinuity. Case 2. $x_0$ is called removal discontinuity. For this case, we can set a new value for $f(x_{0})$ to make continuous.

Continuous functions For a function $f(x)$ is continuous at $x = x_0$ , $f(x_0)$ is defined and limit exists at and their values are equal.

Example 1..27 Find the following function is continuous at $x = 2$

$\displaystyle f(x) = \left\{\begin{array}{rl} x^{2}, & x \neq 2\\ 0, & x = 2 \end{array}\right.$

Check the existence of Left-hand limit and right-hand limit.

SOLUTION Since $\lim_{x \rightarrow 2}f(x) = \lim_{x \rightarrow 2} x^{2} = 4.$ and $f2. = 0$ , we have $\lim_{x \rightarrow 2}f(x) \neq f2.$ . Thus $f(x)$ is discontinuous $x = 2$ . The graph of function has a jump at $x = 2$ $\blacksquare$

Exercise 1..27 Find the following function is continuous at $x = 0$

$\displaystyle f(x) = \left\{\begin{array}{rl} x\sin{\frac{1}{x}}, & x \neq 0\\ 0, & x = 0 \end{array}\right.$

Bibrated function should be squeezed.

SOLUTION Since $\sin{\frac{1}{x}}$ is bibrating, it should be squeezed by the absolute value.

$\displaystyle 0 \leq \vert x \sin{\frac{1}{x}}\vert \leq \vert x\vert.$

Note that $\lim_{x \to 0} 0 = 0$ and $\lim_{x \to 0}\vert x\vert = 0$ . Thus by the squeezing theorem, we have

$\displaystyle \lim_{x \rightarrow 0} \vert x \sin{\frac{1}{x}}\vert = 0.$

Since $\vert\vert f(x)\vert - 0\vert < \varepsilon \Rightarrow \vert f(x) - 0\vert < \varepsilon$ , we have

$\displaystyle \lim_{x \rightarrow 0} x \sin{\frac{1}{x}} = 0.$

Note that

. Thus

is continous at $x = 0$

$\blacksquare$

Continuous on an Interval If $f(x)$ is continuous at every value in I, then we say $f(x)$ is continuous on I.

**Figure 1.43:** Exercise1-27
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$0 \leq \vert\sin{\frac{1}{x}}\vert \leq 1$ is multiplied by $\vert x\vert$ . Then $0 \leq \vert x\sin{\frac{1}{x}}\vert \leq \vert x\vert$ .

Continuity Properties

Theorem 1..9 Suppose that $f(x)$

and

is continuous at $x = x_{0}$ and $c$

is a constant. Then

$1. f(x) \pm g(x)$ is continuous at $x = x_{0}$

$2. cf(x)$ is continuous at $x = x_{0}$

$3. f(x)g(x)$ is continuous at $x = x_{0}$

$4. \displaystyle{\frac{f(x)}{g(x)}}$ is continuous at $x = x_{0}$ provided $g(x_{0}) \neq 0$

Continuity Properties Polynomial, $\sin{x}$ , $\cos{x}$ , $\tan^{-1}{x}$ are continous on $(-\infty,\infty)$ . Rational function is continuous except at the value where the denominator is 0. Continous functions are continuous after the four arithmetic operations.

Proof 1. $\lim_{x \rightarrow x_{0}}f(x) = f(x_{0}), \lim_{x \rightarrow x_{0}}g(x) = g(x_{0})$ , Thus

$\displaystyle \lim_{x \rightarrow x_{0}}(f \pm g)(x) = \lim_{x \rightarrow x_{0... ... \pm \lim_{x \rightarrow x_{0}}g(x) = f(x_{0}) \pm g(x_{0}) = (f \pm g)(x_{0})$

Similarly for 2. , 3. , 4. $\blacksquare$

Composite Continuous Functions

Theorem 1..10 Suppose $y = f(x)$

is continuous at $x = x_{0}$ and $z = g(y)$

is continuous at $y = f(x_{0})$ . Then the composite function $z =g(f(x))$

is Continuous at $x = x_{0}$ .

Proof $\displaystyle{\lim_{x \rightarrow x_{0}}g(f(x)) = g(\lim_{x \rightarrow x_{0}}f(x)) = g(f(x_{0}))}$ $\blacksquare$

As you can see, a continuous function is easy function to use.

Example 1..28 Show the following function is continuous at $x \neq 0$ .

$\displaystyle f(x) = x^{2}\sin{(\frac{1}{x})}$

Use the continuity property.

SOLUTION Note that $\displaystyle{\sin{(\frac{1}{x})}}$ can be thought of a composite function of $\sin{z}$ and $\displaystyle{z = \frac{1}{x}}$ . Now $\sin{z}$ and $\displaystyle{z = \frac{1}{x}}$ are continuous function of $z$ and $x \neq 0$ . Thus $\displaystyle{\sin{(\frac{1}{x})}}$ is continuous at $x \neq 0$ . Also $x^{2}$ is continuous on $(-\infty,\infty)$ . By 1.9, for $x \neq 0$ $\displaystyle{x^{2}\sin{\frac{1}{x}}}$ is continuous $\blacksquare$

Exercise 1..28 Show the following function is continuous on $x \neq 0$ .

$\displaystyle f(x) = \frac{\sin{x}}{e^{\frac{1}{x}}}$

SOLUTION $\sin{x}$ is continuous on $(-\infty,\infty)$ and $e^x$

is continuous on $(-\infty,\infty)$ . Note that $\frac{1}{x}$ is continuous on $x \neq 0$ . Thus, $e^{\frac{1}{x}}$ is continuous on $x \neq 0$ . Finally, a produnct of continuous functions is continuous, we have $\frac{\sin{x}}{e^{\frac{1}{x}}}$ is continuous on $x \neq 0$ $\blacksquare$

**Figure 1.44:** Graph of
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Subsections

Properties of Continuous Functions