Continuity

Defintions and Properties

Definition

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

• \(f(a)\) is defined

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

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

Intuitively, \(f(x)\) is continuous at \(x=a\) if we can draw the graph of \(y=f(x)\) without picking our pencil at \(x=a\).

Definition

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 \(x = a\) if \(g(a)\neq 0\)

Continuity of Polynomial and Rational Functions

Polynomial and rational functions are continuous on their domains.

• Polynomial functions are continuous everywhere.

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

Example 1

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 2

Find the discontinuities

Find the discontinuities of the function \(f(x) = \dfrac{2x^4-2x^3-12x^2}{x^4-9x^2}\).

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 \(x=-3\).

Discontinuities of Rational Functions

Rational functions have discontinuities whenever the denominator is equal to zero. When finding discontinuities of a rational function, do not simplify the rational function by factoring both numerator and denominator and then cancelling common factors. Just factor the denominator and set each factor equal to zero.

Example 3

Choosing a parameter to make function continuous

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

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

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

From the left:

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

From the 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 \(x=2\).