How to Remove Discontinuity from a Function

Understanding Removable Discontinuities

• Definition: A removable discontinuity at a point \( x = a \) occurs if the limit of \( f(x) \) as \( x \) approaches \( a \) exists, but either \( f(a) \) is not defined or \( f(a) \) is not equal to this limit.
• Graphical Representation: On a graph, a removable discontinuity might appear as a ‘hole’ or a ‘gap’ in the function.

Examples of Removable Discontinuities

1. Undefined at the Point:
   Function \( f(x) = \frac{(x^2 – 1)}{(x – 1)} \) is undefined at \( x = 1 \) but the limit as \( x \) approaches \(1\) exists.
2. Defined but not Matching the Limit:
   Function \( g(x) = \begin{cases}
   x + 2, & \text{if } x \neq 3 \\
   7, & \text{if } x = 3
   \end{cases} \) is defined at \( x = 3 \) but does not match the limit as \( x \) approaches \(3\).

Process of Removing Discontinuities

1. Identify the Point of Discontinuity: Determine the point \( x = a \) where the discontinuity exists.
2. Evaluate the Limit: Find the limit of the function as \( x \) approaches the point of discontinuity. Let this limit be \( L \).
3. Redefine the Function: Modify the function’s definition so that \( f(a) = L \). This step effectively ‘fills the hole’ or ‘bridges the gap’ in the function at the discontinuity.

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Redefining Functions with Removable Discontinuities

• For Undefined Points: If \( f(a) \) was originally undefined, redefine ( f(a) ) to be the limit value. For example, redefine \( f(x) = \frac{(x^2 – 1)}{(x – 1)} \) at \( x = 1 \) to be \( f(1) = 2 \).
• For Defined but Inconsistent Points: If \( f(a) \) was defined but different from the limit, adjust the definition to match the limit. In the case of \( g(x) \), redefine \( g(3) \) to be \( 5 \).

Theoretical and Practical Implications

• Continuous Extension: The modified function, now continuous at the point of discontinuity, is often referred to as a “continuous extension” of the original function.
• Mathematical Analysis: This process is significant in mathematical analysis for simplifying functions and making sure theorems that need continuity, like the Intermediate Value Theorem, apply.
• Real-world Applications: In real-world scenarios, such as signal processing or data interpolation, removing discontinuities can be fundamental for creating smooth, predictable models.

Step-by-Step: Removing a Discontinuity

1. Find the trouble point. Look for an x-value where the function is undefined, such as a zero in the denominator, or where a piecewise function changes rules.
2. Find the limit there. For a rational function, factor the top and bottom, cancel the common factor, and substitute.
3. Check that it is removable. The limit must exist as a finite number, with both sides approaching the same value.
4. Redefine the function at that point so its value equals the limit.

Example 1: f(x) = (x² + x − 6)/(x − 2). Factor: (x + 3)(x − 2)/(x − 2) = x + 3 for x ≠ 2. The limit as x → 2 is 5. The redefined function is f(x) = (x² + x − 6)/(x − 2) for x ≠ 2, and f(2) = 5.

Example 2 (wrong value at the point): g(x) = x² for x ≠ 3, and g(3) = 4. The limit as x → 3 is 9, but g(3) = 4. Change the value to g(3) = 9, and g is continuous.

Finding k to Make a Function Continuous

Many problems hide the value you need behind a letter.

Example 3: f(x) = (x² − 25)/(x − 5) for x ≠ 5, and f(5) = k. Simplify to x + 5, so the limit at 5 is 10. Set k = 10.

Example 4: f(x) = ax + 1 for x < 2, and f(x) = x² + 3 for x ≥ 2. The right side gives 2² + 3 = 7 at x = 2. The left side approaches 2a + 1. Set 2a + 1 = 7, so a = 3.

Common mistake: Canceling terms instead of factors. In (x + 4)/(x + 2), you cannot cancel the x’s. Only a common factor such as (x − 2) can be canceled.

FAQ

What is a removable discontinuity in a function?

A removable discontinuity happens at x = a when the limit of f(x) as x approaches a exists, but f(a) is either undefined or not equal to that limit. On a graph it shows up as a single hole. For example, f(x) = (x² − 4)/(x − 2) is undefined at x = 2, but the values approach 4, so the hole sits at (2, 4).

How do you remove a discontinuity from a function?

Find the limit at the trouble point, then redefine the function so its value there equals the limit. Take f(x) = (x² + x − 6)/(x − 2). Factor the top as (x + 3)(x − 2), cancel x − 2, and the limit as x approaches 2 is 2 + 3 = 5. Define f(2) = 5 and the new function is continuous at x = 2.

How do you find a removable discontinuity of a function?

For a rational function, factor the numerator and denominator completely. A factor that cancels from both gives a removable discontinuity. In f(x) = (x² − 5x + 6)/(x² − 9), the top is (x − 2)(x − 3) and the bottom is (x − 3)(x + 3). The factor x − 3 cancels, so there is a hole at x = 3, at height (3 − 2)/(3 + 3) = 1/6.

How do you find a non-removable discontinuity of a function?

Look for points where the limit does not exist. A denominator factor that does not cancel creates a vertical asymptote, such as x = −3 in (x − 2)/(x + 3). A piecewise function whose left and right pieces meet at different heights has a jump. In both cases no single value can fill the gap, so the discontinuity cannot be removed.

Is a removable discontinuity the same as a hole?

Yes, in most cases. A removable discontinuity is drawn as a hole, an open circle on the curve, because only one point is missing or out of place. Sometimes the function is defined at that x-value but at the wrong height, so the graph shows an open circle on the curve and a separate dot above or below it. Both are removable.

Is a removable discontinuity undefined?

Not always. Often the function is undefined at that point, as with (x² − 1)/(x − 1) at x = 1. But a function can also be defined there with the wrong value. If h(x) = x + 1 for x ≠ 2 and h(2) = 7, the limit at 2 is 3, not 7, so there is a removable discontinuity even though h(2) exists.

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How do you write the redefined function after removing a discontinuity?

Write it as a piecewise function. Keep the original rule for every x except the trouble point, and add a second line giving the limit value at that point. For (x² − 9)/(x − 3), write f(x) = (x² − 9)/(x − 3) when x ≠ 3, and f(x) = 6 when x = 3. You can also simply write f(x) = x + 3.

How do you find the value of k that makes a function continuous?

Find the limit of the function at the point where k appears, then set k equal to it. If the function is piecewise, make the left-hand and right-hand values match at the boundary and solve for k. For f(x) = kx for x < 1 and x + 4 for x ≥ 1, set k = 5.

What is the difference between a hole and a vertical asymptote?

Both can come from a zero in the denominator. If the factor cancels with a factor in the numerator, you get a hole, which is removable. If it does not cancel, the function grows without bound there, and you get a vertical asymptote, which is not removable. In (x − 2)/((x − 2)(x + 1)), there is a hole at x = 2 and an asymptote at x = −1.

Does removing a discontinuity change the graph?

Only at one point. The graph before has an open circle (a hole) at the trouble point, or a dot in the wrong place. After you redefine the function, that point is filled in on the curve. Everything else stays exactly the same.