Continuity

Defintion and Properties

Definition

A function \(f(x)\) is continuous at a number \(x = a\) if the following conditions are satisfied:

• \(f(a)\) is defined

• \(\displaystyle{\lim_{x \to \\a}} f(x)\) exists (rule for existence: \(\displaystyle{\lim_{x \to \\a^-}} f(x) = \displaystyle{\lim_{x \to \\a^+}} f(x)\) )

• \(\displaystyle{\lim_{x \to \\a}} f(x) = f(a)\)

The function \(f(x)\) is continuous on the interval \((a, b)\) if it is continuous at each point in the interval.

If \(f(x)\) is not continuous at \(x=a\), then \(f(x)\) is said to be discontinuous at \(x=a\).

Properties of Continuous Functions

If \(f(x)\) and \(g(x)\) are continuous at \(x = a\) then:

• \([f(x)]^n\) where \(n\) is a real number, is continuous at \(x=a\) if \([f(a)]^n\) is defined.

• \(f(x) \pm g(x)\) is continuous at \(x = a\)

• \(f(x) \times g(x)\) is continuous at \(x = a\)

• \(\dfrac{f(x)}{g(x)}\) is continuous at \(x = a\) provided \(g(a)\neq 0\)

Continuity of Polynomial and Rational Functions

Polynomial and rational functions are continuous on their domains.

• A polynomial function \(y = p(x)\) is continuous everywhere.

• A rational function \(R(x) = \dfrac{p(x)}{q(x)}\) is continuous everywhere \(q(x) \neq 0\).

Examples

Example: Continuity at a point

Is the following function continuous at \(x=1\)?

\[\begin{split} f(x) = \begin{cases} -x^3+x^2-1 & \hbox{if } x\leq 1 \\ 3x^2-x-3 & \hbox{if } x > 1 \end{cases} \end{split}\]

Step 1:   Determine if   \(f(x)\)   is defined at   \(x=1\).

When \(x=1\), \(f(x)\) is defined by \(f(x) = -x^3+x^2-1\). Therefore,

\[\begin{align*} f(1) &= -(1)^3+(1)^2-1 & \text{plug in $x=1$}\\ &= -1 \end{align*}\]

Since \(f(x)\) is defined at \(x=1\), we now check to see if the limit exists.

Step 2:   Determine if the limit at   \(x=1\)   exists.

For the left-hand limit, \(x\to1^-\) means \(x< 1\), and therefore \(f(x) = -x^3+x^2-1\).

\[\begin{align*} \lim_{x\to 1^-} f(x) &= \lim_{x\to 1^-} -x^3+x^2-1\\ &= -1^3+1^2-1 & \text{plug in $x=1$}\\ &= -1 \end{align*}\]

For the right-hand limit, \(x\to1^+\) means \(x> 1\), and therefore \(f(x) = 3x^2-x-3\).

\[\begin{align*} \lim_{x\to 1^+} f(x) &= \lim_{x\to 1^+} 3x^2-x-3 \\ &= 3(1)^2-1-3 & \text{plug in $x=1$} \\ &= -1 \end{align*}\]

Since both one-sided limits exist and are equal to \(-1\), we conclude that

\[\lim_{x\to 1} f(x) = -1\]

Since the limit exists, we now check to see if the value of the function and the limit are equal to each other.

Step 3:   Compare   \(f(1)\)   and   \(\lim\limits_{x \to 1} f(x)\).

Both the function at \(x=1\) and the limit of the function as \(x\) approaches \(1\) are equal to \(-1\).

\[f(1) = -1 = \lim_{x\to 1}f(x)\]

Step 4:   Conclusion

Since \(f(1)\) is defined, \(\lim\limits_{x \to 1} f(x)\) exists, and \(\lim\limits_{x \to 1}f(x) = f(1)\), we conclude that \(f(x)\) is continuous at \(x=1\).

Example: Discontinuities

Find the discontinuities of \(f(x)\) where

\[f(x) = \frac{2x^4-2x^3-12x^2}{x^4-9x^2}.\]

Step 1:   Factor the denominator.

\[\begin{align*} x^4-9x^2 &= x^2(x^2-9) && \text{pull out common factor of $x^2$}\\ &= x^2(x-3)(x+3) && \text{since $A^2 - B^2 = (A-B)(A+B)$} \end{align*}\]

Step 2:   Set each factor of the denominator equal to zero.

\[\begin{align*} x^2 &= 0 ~~~~\hbox{ when $x=0$}\\ x-3 &= 0 ~~~~\hbox{ when $x = 3$}\\ x+3 &= 0 ~~~~\hbox{ when $x = -3$} \end{align*}\]

Therefore, \(f(x)\) has discontinuities at \(x=0\), \(x=3\), and \(x=-3\).

Warning

Remember, rational functions have discontinuities whenever the denominator is equal to zero. There is no need to factor and/or simplify the numerator when finding discontinuities of a rational function.

Example: Choosing a parameter to make a function continuous

Find the value of \(k\) that makes \(f(x)\) continuous at \(x=2\).

\[\begin{split} f(x) = \begin{cases} x+k & \hbox{ if $x < 2$}\\ x^2 + kx - 7 & \hbox{ if $x \geq 2$} \end{cases} \end{split}\]

Step 1:   Evaluate   \(\displaystyle{\lim_{x \to 2}} f(x)\)   from left and right.

Left

\[\begin{align*} \lim_{x \to 2^-} f(x) &= \lim_{x \to 2^-} x+k \\ &=2+k \end{align*}\]

Right

\[\begin{align*} \lim_{x \to 2^+} f(x) &= \lim_{x \to 2^+} x^2 + kx -7 \\ &=2^2+k(2)-7 \\ &=2k-3 \end{align*}\]

Step 2:   Set the left and right limits equal to each other.

\[2+k = 2k-3\]

Solve for \(k\) by adding 3 to both sides and subtracting \(k\) from both sides. Thus \(k=5\) makes the function \(f(x)\) continuous at \(x=2\).