# Continuity of Functions

## Solved Problems

Click or tap a problem to see the solution.

### Example 3

Using Cauchy definition, prove that $\lim\limits_{x \to 4} \sqrt x = 2.$

### Example 4

Show that the cubic equation $2{x^3} - 3{x^2} - 15 = 0$ has a solution in the interval $$\left( {2,3} \right)$$.

### Example 5

Show that the equation ${x^{1000}} + 1000x - 1 = 0$ has a root.

### Example 6

Let

${f\left(x \right) = \begin{cases} x^2 + 2, & x \lt 0 \\ ax + b, & 0 \le x \lt 1 \\ 3 + 2x - {x^2}, & x \ge 1 \end{cases}}.$

Determine $$a$$ and $$b$$ so that the function $$f\left(x \right)$$ is continuous everywhere.

### Example 7

If the function

${f\left(x \right) = \begin{cases} \cos \left( {2\pi x- a} \right), &x \lt -1 \\ x^3 + 1, &x \ge -1 \end{cases}}$

is continuous, what is the value of $$a?$$

### Example 3.

Using Cauchy definition, prove that $\lim\limits_{x \to 4} \sqrt x = 2.$

Solution.

Let $$\varepsilon \gt 0$$. We must find some number $$\delta \gt 0$$ so that for all $$x$$ with

$\left| {x - 4} \right| \lt \delta ,$

the following inequality is valid:

$\left| {f\left( x \right) - f\left( 4 \right)} \right| \lt \varepsilon \;\;\text{or}\;\;\left| {\sqrt x - 2} \right| \lt \varepsilon .$

We can write the last expression as

$- \varepsilon \lt \sqrt x - 2 \lt \varepsilon ,\;\; \Rightarrow 2 - \varepsilon \lt \sqrt x \lt \varepsilon + 2,\;\; \Rightarrow \left( {2 - \varepsilon } \right)^2 \lt x \lt {\left( {\varepsilon + 2} \right)^2},\;\; \Rightarrow {\varepsilon ^2} - 4\varepsilon + 4 \lt x \lt {\varepsilon ^2} + 4\varepsilon + 4.$

Hence,

${\varepsilon ^2} - 4\varepsilon \lt x - 4 \lt {\varepsilon ^2} + 4\varepsilon .$

Notice that our function takes only non-negative values. Therefore the $$\varepsilon$$-neighborhood at the given point must satisfy the condition $$\varepsilon \le 2$$. In this case the left side $${\varepsilon ^2} - 4\varepsilon$$ of the inequality will be negative. It follows from here that

${\varepsilon ^2} - 4\varepsilon \lt x - 4 \lt 4\varepsilon - {\varepsilon ^2}\;\;\text{or}\;\;\left| {x - 4} \right| \lt 4\varepsilon - {\varepsilon ^2}.$

Thus, if we take $$\delta \le {\varepsilon ^2} - 4\varepsilon$$, then for all $$x$$ with $$\left| {x - 4} \right| \lt \delta$$, we will have $$\left| {f\left( x \right)} - 2 \right| \lt \varepsilon$$. For example, if $$\varepsilon = 0.1$$, then the value of $$\delta$$ must be $$\delta \le {\varepsilon ^2} - 4\varepsilon$$ $$= 0.4 - 0.01$$ $$= 0.39.\,$$ This means, by Cauchy definition, that

$\lim\limits_{x \to 4} \sqrt x = 2.$

### Example 4.

Show that the cubic equation $2{x^3} - 3{x^2} - 15 = 0$ has a solution in the interval $$\left( {2,3} \right)$$.

Solution.

Let $$f\left( x \right) = 2{x^3} - 3{x^2} - 15$$. Calculate the values of the function at $$x = 2$$ and $$x = 3.$$

$f\left( 2 \right) = 2 \cdot {2^3} - 3 \cdot {2^2} - 15 = - 11,\;\;f\left( 3 \right) = 2 \cdot {3^3} - 3 \cdot {3^2} - 15 = 12.$

Hence, we have $$f\left( 2 \right) \lt 0$$ and $$f\left( 3 \right) \gt 0$$, or

$f\left( 2 \right) \lt 0 \lt f\left( 3 \right).$

By the intermediate value theorem, this means that there exists a number $$c$$ in the interval $$\left( {2,3} \right)$$ such that $$f\left( c \right) = 0.$$

Thus, the equation has a solution in the interval $$\left( {2,3} \right).$$

### Example 5.

Show that the equation ${x^{1000}} + 1000x - 1 = 0$ has a root.

Solution.

As the function $$f\left( x \right) = {x^{1000}} + 1000x - 1$$ is a polynomial, it is continuous. We notice that

$f\left( 0 \right) = {0^{1000}} + 1000 \cdot 0 - 1 = - 1,\;\;f\left( 1 \right) = {1^{1000}} + 1000 \cdot 1 - 1 = 1000.$

Therefore $$f\left( 0 \right) \lt 0 \lt f\left( 1 \right)$$. Then we can conclude by the intermediate value theorem, that there exists a number $$c$$ in the interval $$\left( {0,1} \right)$$ such that $$f\left( c \right) = 0$$. Thus, the equation has a solution in the interval $$\left( {0,1} \right).$$

### Example 6.

Let

${f\left(x \right) = \begin{cases} x^2 + 2, & x \lt 0 \\ ax + b, & 0 \le x \lt 1 \\ 3 + 2x - {x^2}, & x \ge 1 \end{cases}}.$

Determine $$a$$ and $$b$$ so that the function $$f\left(x \right)$$ is continuous everywhere.

Solution.

The left-side limit at $$x = 0$$ is

$\lim\limits_{x \to 0 - 0} f\left( x \right) = \lim\limits_{x \to 0 - 0} \left( {{x^2} + 2} \right) = 2.$

Then the value of $$ax + b$$ at $$x = 0$$ must be equal to $$2.$$

$ax + b = 2,\;\; \Rightarrow a \cdot 0 + b = 2,\;\; \Rightarrow b = 2.$

Similarly, the right-side limit at $$x = 1$$ is

$\lim\limits_{x \to 1 + 0} f\left( x \right) = \lim\limits_{x \to 1 + 0} \left( {3 + 2x - {x^2}} \right) = 3 + 2 - 1 = 4.$

As you can see, the value of $$ax + 2$$ at $$x = 1$$ must be equal to $$4.$$

$ax + 2 = 4,\;\; \Rightarrow a \cdot 1 + 2 = 4,\;\; \Rightarrow a = 2.$

For given values of $$a$$ and $$b$$, the function $$f\left(x \right)$$ is continuous. The graph of the function is sketched in Figure $$4.$$

### Example 7.

If the function

${f\left(x \right) = \begin{cases} \cos \left( {2\pi x- a} \right), &x \lt -1 \\ x^3 + 1, &x \ge -1 \end{cases}}$

is continuous, what is the value of $$a?$$

Solution.

We calculate the left-hand and right-hand limits at $$x = -1.$$

$\lim\limits_{x \to - 1 - 0} f\left( x \right) = \lim\limits_{x \to - 1 - 0} \cos \left( {2\pi x - a} \right) = \cos \left( { - 2\pi - a} \right) = \cos a,$
$\lim\limits_{x \to - 1 + 0} f\left( x \right) = \lim\limits_{x \to - 1 + 0} \left( {{x^3} + 1} \right) = \left( { - 1} \right)^3 + 1 = 0.$

The function will be continuous at $$x = -1,$$ if

$\lim\limits_{x \to - 1 - 0} f\left( x \right) = \lim\limits_{x \to - 1 + 0} f\left( x \right)\;\;\text{or}\;\;\cos a = 0.$

Hence,

$a = \frac{\pi }{2} + \pi n,\;\;n \in \mathbb{Z}.$