Calculus

Limits and Continuity of Functions

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Continuity of Functions

Solved Problems

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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 \(4.\)

Continuity of function
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,

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