MA005: Calculus I

When $x$ is close to 1, the values of $f(x)$ are close to the value $f(1)$ and the graph of $\mathrm{f}$ in Fig. 1 does not have a hole or break at $\mathrm{x}=1$. The graph of $\mathrm{f}$ is connected at $\mathrm{x}=1$ and can be drawn without lifting your pencil. At $\mathrm{x}=2$ and $\mathrm{x}=4$ the graph of $\mathrm{f}$ has holes, and at $\mathrm{x}=3$ the graph has a break. The function $\mathrm{f}$ is also continuous at $1.7$ (why?), and at every point shown except at 2, 3, and 4.

Sometimes the definition of continuous (the substitution condition for limits) is easier to use if we break it into several smaller pieces and then check whether or not our function satisfies each piece.

$\{\mathrm{f}$ is continuous at $\mathrm{a}$ $\}$ if and only if $\left\{\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})\right\}$ if and only if

If $\mathrm{f}$ satisfies conditions (i), (ii) and (iii), then $\mathrm{f}$ is continuous at $\mathrm{a}$. If $\mathrm{f}$ does not satisfy one or more of the three conditionsat $\mathrm{a}$, then $f$ is not continuous at $a$.

For the function in Fig. 1, at $a=1$, all 3 conditions are satisfied, and $f$ is continuous at $1 .$ At $a=2$ conditions (i) and (ii) are satisfied but not (iii), so $\mathrm{f}$ is not continuous at $2 .$ At $\mathrm{a}=3$, condition (i) is satisfied but (ii) is violated, so $f$ is not continuous at $3 .$ At $a=4$, condition (i) is violated, so $f$ is not continuous at 4.

A function is continuous on an interval if it is continuous at every point in the interval. A function $\mathrm{f}$ is continuous from the left at $\mathrm{a}$ if $\lim\limits_{x \rightarrow a^{-}} \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})$, and is continuous from the right if $\lim\limits_{x \rightarrow a^{+}} f(x)=f(a)$.

Example 1: Is $f(x) = \left\{ \begin{array}{lll}x + 1 & \mbox {if x ≤ 1 } \\ 2 & \mbox { if 1 < x ≤ 2 } \\ 1/(x–3) & \mbox { if x > 2} \end{array} \right.$ continuous at $1,2,3$?

Solution: We could answer these questions by examining the graph of $\mathrm{f}(\mathrm{x})$, but lets try them without the graph. At $\mathrm{a}=\mathbf{1}, \mathrm{f}(\mathbf{1})=2$ and the left and right limits are equal, $\mathrm{f}(\mathrm{x})=\lim\limits_{x \rightarrow 1^{-}}(x+1)=2$ and $\lim\limits_{x \rightarrow 1^{+}} \mathrm{f}(\mathrm{x})=\lim\limits_{x \rightarrow 1^{+}} 2=2$, so $\mathrm{f}$ is continuous at $\mathbf{1}$

At $\mathrm{a}=\mathbf{2}, \mathrm{f}(\mathbf{2})=2$, but the left and right limits are not equal,

$\lim\limits_{x \rightarrow 2^{-}} f(x)=\lim\limits_{x \rightarrow 2^{-}} 2=2$ and $\lim\limits_{x \rightarrow 2^{+}} f(x)=\lim\limits_{x \rightarrow 2^{+}} 1 /(x-3)=-1$, so $\mathrm{f}$ fails condition (ii) and is not continuous at 2. $\mathrm{f}$ is continuous from the left at $\mathbf{2}$, but not from the right.

At $\mathrm{a}=\mathbf{3}, \mathrm{f}(\mathbf{3})=1 / 0$ which is undefined so $\mathrm{f}$ is not continuous at $\mathbf{3}$ because it fails condition (i).

Example 2: Where is $f(x)=3 x^{2}-2 x$ continuous?

Solution: At every point. By the Substitution Theorem for Polynomials, every polynomial is continuous everywhere.

Example 3: Where are $\mathrm{g}(\mathrm{x})=\frac{\mathrm{x}+5}{\mathrm{x}-3} \quad$ and $\mathrm{h}(\mathrm{x})=\frac{\mathrm{x}^{2}+4 \mathrm{x}-5}{\mathrm{x}^{2}-4 \mathrm{x}+3}$ continuous?

Solution: $\mathrm{g}$ is a rational function so by the Substitution Theorem for Polynomials and Rational Functions it is continuous everywhere except where its denominator is $0: \mathrm{g}$ is continuous everywhere except $3 .$ The graph of g (Fig. 2) is connected everywhere except at 3 where it has a vertical asymptote. $\mathrm{h}(\mathrm{x})=\frac{(\mathrm{x}-1)(\mathrm{x}+5)}{(\mathrm{x}-1)(\mathrm{x}-3)}$ is also continuous everywhere except where its denominator is $0: \mathrm{h}$ is continuous everywhere except 3 and $1 .$ The graph of $h$ (Fig.3) is connected everywhere except at 3 where it has a vertical asymptote and at 1 where it has a hole: $\mathrm{f}(1)=0 / 0$ is undefined.

Example 4: $\quad$ Where is $f(x)=\operatorname{INT}(x)$ continuous?

Solution: The graph of $\mathrm{y}=\operatorname{INT}(\mathrm{x})$ seems to be connected except at each integer, and at each integer there is a "jump" (Fig.4).

If $\mathrm{a}$ is an integer, then $\lim\limits_{x \rightarrow a^{-}} \operatorname{INT}(x)=a-1$, and $\lim\limits_{x \rightarrow a^{+}} \operatorname{INT}(x)=a$, so $\lim\limits_{x \rightarrow a} \operatorname{INT}(x)$ is undefined, and $\mathrm{INT(x)}$ is not continuous.

If $\mathrm{a}$ is not an integer, then the left and right limits of $\mathrm{INT(x)}$, as $x \rightarrow a$, both equal $\mathrm{INT(a)}$ so

$\lim\limits_{x \rightarrow a} \operatorname{INT}(x)=\operatorname{INT}(a)=f(a)$ and $\operatorname{INT}(x)$ is continuous. $f(x)=\operatorname{INT}(x)$ is continuous except at the integers.

In fact, $f(x)=\operatorname{INT}(x)$ is continuous from the right everywhere and is continuous from the left everywhere except at the integers.

Practice 1: $\quad$ Where is $\mathrm{f}(\mathrm{x})=\mid \mathrm{xl} / \mathrm{x}$ continuous?