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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 $f(x)$ is continous at $x = x_{0}$.
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.

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

Figure 1.5: Discontinuous

\includegraphics[width=10cm]{SOFTFIG-1/Fig1-5-1.eps}

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.

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

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

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 $x = 0$.

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

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 $f(0) = 0$. Thus $f(x)$ 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.

Continuity Properties

Theorem 1..9   Suppose that $f(x)$ and $g(x)$ 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 $g(x_{0}) \neq 0$

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 $x \neq 0$.

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

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 $x \neq 0$ $\ \blacksquare$
Subsections
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