2.4: Continuity
- Page ID
- 145070
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Continuity
A function that is "friendly" and doesn’t have any breaks or jumps in it is called continuous. More formally,
A function \(\bf f\) is continuous at \(\bf x = a \) if and only if \( \lim\limits_{x\to a} \mathbf{f(x)} = \mathbf{f(a)}\).
The graph below illustrates some of the different ways a function can behave at and near a point, and the table contains some numerical information about the function and its behavior.
| \( a \) | \( f(a) \) | \( \lim\limits_{x\to a} f(x) \) |
| 1 | 2 | 2 |
| 2 | 1 | 2 |
| 3 | 2 | Does not exist (DNE) |
| 4 | Undefined | 2 |
Based on the information in the table, we can conclude that \(f\) is continuous at 1 since \( \lim\limits_{x\to 1} f(x) = 2 = f(1)\).
We can also conclude from the information in the table that \(f\) is not continuous at 2 or 3 or 4, because \( \lim\limits_{x\to 2} f(x) \neq f(2) \), \( \lim\limits_{x\to 3} f(x) \neq f(3) \), and \( \lim\limits_{x\to 4} f(x) \neq f(4) \).
The behaviors at \(x = 2\) and \(x = 4\) exhibit a hole in the graph, sometimes called a removable discontinuity, since the graph could be made continuous by changing the value of a single point. The behavior at \( x = 3 \) is called a jump discontinuity, since the graph jumps between two values.
So which functions are continuous? It turns out pretty much every function you’ve studied is continuous where it is defined: polynomial, radical, rational, exponential, and logarithmic functions are all continuous where they are defined. Moreover, any combination of continuous functions is also continuous.
This is helpful, because the definition of continuity says that for a continuous function, \( \lim\limits_{x\to a} f(x) = f(a) \). That means for a continuous function, we can find the limit by direct substitution (evaluating the function) if the function is continuous at \(a\).
Evaluate using continuity, if possible:
- \( \lim\limits_{x\to 2} x^3-4x \)
- \( \lim\limits_{x\to 2} \dfrac{x-4}{x+3} \)
- \( \lim\limits_{x\to 2} \dfrac{x-4}{x-2} \)
Solution
- The given function is polynomial, and is defined for all values of \(x\), so we can find the limit by direct substitution:\[ \lim\limits_{x\to 2} x^3-4x = 2^3-4(2) = 0. \nonumber \]
- The given function is rational. It is not defined at \(x = -3\), but we are taking the limit as \(x\) approaches 2, and the function is defined at that point, so we can use direct substitution:\[ \lim\limits_{x\to 2} \dfrac{x-4}{x+3} = \dfrac{2-4}{2+3}= -\dfrac{2}{5}. \nonumber \]
- This function is not defined at \(x = 2\), and so is not continuous at \(x = 2\). We cannot use direct substitution.