How Would You Remove The Discontinuity Of F

A discontinuity in a function $f(x)$ represents a break, gap, or hole in its graph where the function fails to be continuous. For students and professionals working with calculus and real analysis, understanding how to remove the discontinuity of f is a fundamental skill. It transforms a "broken" function into a smooth, predictable tool suitable for differentiation, integration, and modeling real-world phenomena. The process relies entirely on classifying the type of discontinuity present, as the method for removal differs significantly between a simple hole and an infinite asymptote.

Understanding the Definition of Continuity

Before attempting to fix a break, one must verify exactly why the function fails the continuity test at a specific point $x = c$. A function $f$ is continuous at $c$ if and only if three conditions are met simultaneously:

1. $f(c)$ is defined: The function must have an actual output value at $c$.
2. $\lim_{x \to c} f(x)$ exists: The left-hand limit and right-hand limit must be equal and finite.
3. Equality holds: $\lim_{x \to c} f(x) = f(c)$.

If any of these conditions fail, a discontinuity exists. The strategy for removal targets the specific condition that is violated.

Classifying Discontinuities: The Roadmap to Repair

Not all discontinuities are created equal. Mathematicians categorize them into three main types, and identifying the category is the first step in the removal process Not complicated — just consistent..

1. Removable Discontinuity (The "Hole")

This is the only type that can be truly "removed" to make the function continuous at that specific point without changing the function's formula elsewhere. It occurs when the limit exists (Condition 2 holds), but either $f(c)$ is undefined (Condition 1 fails) or $f(c) \neq \lim_{x \to c} f(x)$ (Condition 3 fails). Graphically, this looks like a single missing point—a hole—in an otherwise connected curve.

2. Jump Discontinuity

Here, the left-hand limit $\lim_{x \to c^-} f(x)$ and the right-hand limit $\lim_{x \to c^+} f(x)$ both exist but are not equal. Because the two-sided limit does not exist, Condition 2 fails fundamentally. You cannot "remove" a jump discontinuity at $x=c$ by simply redefining a single point; the function has two different destinations as it approaches $c$ Most people skip this — try not to..

3. Infinite (Essential) Discontinuity

This occurs when the function grows without bound (approaches $\pm \infty$) as $x$ approaches $c$. The limit does not exist in the finite sense. Vertical asymptotes are the hallmark of this type. Like jump discontinuities, these cannot be removed by redefining $f(c)$ because no finite limit exists to assign.

Step-by-Step: Removing a Removable Discontinuity

Since only removable discontinuities can be fixed by redefinition, this is the core procedure implied by the question "how would you remove the discontinuity of f."

Step 1: Identify the Suspect Point

Locate the $x$-value ($c$) where the function is not continuous. This is often where a denominator equals zero (in rational functions), where a piecewise definition changes, or where a square root argument becomes negative.

Step 2: Evaluate the Limit

Calculate the two-sided limit $\lim_{x \to c} f(x)$. Use algebraic manipulation—factoring, rationalizing conjugates, or using trigonometric identities—to resolve the indeterminate form (usually $0/0$).

• If the limit is finite ($L$): Proceed to Step 3. The discontinuity is removable.
• If the limit is infinite or DNE (different one-sided limits): The discontinuity is not removable at $x=c$. Stop here.

Step 3: Redefine the Function

Create a new function, often denoted as $g(x)$ or $\tilde{f}(x)$, defined piecewise: $ g(x) = \begin{cases} f(x) & \text{if } x \neq c \ L & \text{if } x = c \end{cases} $ By explicitly setting the output at $x=c$ equal to the limit $L$, you satisfy all three continuity conditions. The "hole" is plugged.

Worked Examples: Algebraic Techniques in Action

Theory is best understood through practice. Below are the most common algebraic scenarios requiring discontinuity removal.

Example 1: Factoring Rational Functions (The Classic $0/0$)

Consider $f(x) = \frac{x^2 - 4}{x - 2}$.

• Problem: At $x=2$, the denominator is zero. $f(2)$ is undefined.
• Limit Evaluation: Factor the numerator: $\frac{(x-2)(x+2)}{x-2}$.
• Simplify: For $x \neq 2$, cancel the common factor: $x + 2$.
• Find Limit: $\lim_{x \to 2} (x + 2) = 4$.
• Removal: Define the continuous extension $g(x)$: $ g(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & x \neq 2 \ 4 & x = 2 \end{cases} $ Note: $g(x)$ is effectively the line $y = x + 2$ for all real numbers.

Example 2: Rationalizing Numerators/Denominators (Radicals)

Consider $f(x) = \frac{\sqrt{x+4} - 2}{x}$.

• Problem: At $x=0$, we get $\frac{0}{0}$.
• Limit Evaluation: Multiply numerator and denominator by the conjugate $\sqrt{x+4} + 2$: $ \frac{(\sqrt{x+4} - 2)(\sqrt{x+4} + 2)}{x(\sqrt{x+4} + 2)} = \frac{(x+4) - 4}{x(\sqrt{x+4} + 2)} = \frac{x}{x(\sqrt{x+4} + 2)} $
• Simplify: Cancel $x$ (valid for $x \neq 0$): $\frac{1}{\sqrt{x+4} + 2}$.
• Find Limit: $\lim_{x \to 0} \frac{1}{\sqrt{x+4} + 2} = \frac{1}{4}$.
• Removal: Define $f(0) = \frac{1}{4}$.

Example 3: Trigonometric Limits (The $\sin x / x$ Standard)

Consider $f(x) = \frac{\sin(3x)}{x}$ That alone is useful..

• Problem: Undefined at $x=0$.
• Limit Evaluation: Manipulate to use $\lim_{u \to 0} \frac{\sin u}{u} = 1$. $ \frac{\sin(3x)}{x} = 3 \cdot \frac{\sin(3x)}{3x} $ Let $u = 3x$. As $x \to 0, u \to 0$. $ \lim_{x \to 0} 3 \cdot \frac{\sin(3x)}{3x} = 3 \cdot 1 = 3 $
• Removal: Define $f(0) = 3$.

Example 4: Piecewise Function Mismatch

Consider $f(x) = \begin{cases} x^2 + 1 & x < 1 \ 5 & x = 1 \ 2x & x > 1 \end{cases}$

• Check Limits:
  • Left limit: $\lim_{x \to 1^-} (x^2 + 1) = 2$.

Example 4 (continued)

The right‑hand limit is obtained from the branch that applies when (x>1):

[ \lim_{x\to 1^{+}} 2x = 2\cdot 1 = 2 . ]

Since the left‑hand and right‑hand limits coincide (both equal 2), the overall limit exists and

[ \lim_{x\to 1} f(x)=2 . ]

The actual value of the function at the point of interest is (f(1)=5), which differs from the limit. Because the limit exists and is finite, the discontinuity can be eliminated by redefining the function at (x=1) That's the part that actually makes a difference..

Define the continuous extension

[ g(x)=\begin{cases} x^{2}+1, & x<1,\[2pt] 2, & x=1,\[2pt] 2x, & x>1 . \end{cases} ]

Now (g(1)=2) matches the limit, and all three continuity conditions are satisfied. The “hole’’ at (x=1) has been plugged.

Example 5 – Removable Discontinuity in a Quotient with a Repeated Factor

Consider

[ h(x)=\frac{x^{2}-1}{x-1}. ]

Direct substitution gives (\frac{0}{0}), so the function is undefined at (x=1).

Factor the numerator:

[ \frac{(x-1)(x+1)}{x-1}=x+1\qquad (x\neq 1). ]

The limit as (x) approaches 1 is therefore

[ \lim_{x\to 1} h(x)=\lim_{x\to 1}(x+1)=2 . ]

Redefine the function at the problematic point:

[ \tilde{h}(x)=\begin{cases} \dfrac{x^{2}-1}{x-1}, & x\neq 1,\[4pt] 2, & x=1 . \end{cases} ]

The new function (\tilde{h}) is continuous everywhere, including at (x=1) Nothing fancy..

Example 6 – When a Removable Discontinuity Is Not Possible

Take

[ p(x)=\frac{1}{x-2}. ]

As (x\to 2^{-}), the values decrease without bound, while as (x\to 2^{+}) they increase without bound. Consequently

[ \lim_{x\to 2} p(x)\ \text{does not exist (the one‑sided limits are infinite)}. ]

Because the limit is not finite, the discontinuity cannot be removed by any redefinition. The point (x=2) remains a non‑removable (essential) discontinuity.

Summary and Conclusion

To resolve a removable discontinuity, follow these concise steps:

1. Identify the point (c) where the function is undefined or appears inconsistent.
2. Compute the limit (\displaystyle L=\lim_{x\to c}f(x)) using algebraic manipulation, rationalization, trigonometric identities, or known limits. Verify that the left‑hand and right‑hand limits agree and are finite.
3. Redefine the function at (c) so that its value equals the computed limit, producing a piecewise‑defined function that satisfies continuity everywhere.

When the limit fails to exist or diverges, the discontinuity is inherently non‑removable, and no amount of redefinition can restore continuity at that point Easy to understand, harder to ignore..

By mastering the techniques illustrated—factoring, conjugates, standard trigonometric limits, and careful handling of piecewise definitions—students gain a reliable toolkit for tackling the most frequent algebraic obstacles that arise in continuity analysis. This systematic approach not only smooths out “holes’’ in graphs but also reinforces a deeper understanding of how limits govern the behavior of functions.