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1.13 Removing Discontinuities

Keywords

Removing Discontinuities 消除间断点 • Extended Function 扩展函数

Point Redefinition 点重定义 • Piecewise Boundary 分段边界

The Concept of "Repairing" a Function

If a function \(f(x)\) has a removable discontinuity at \(x=c\), it means the limit exists (\(\lim\limits_{x \to c} f(x) = L\)), but the function is either undefined or has a different value at that point. We can "remove" this discontinuity by defining a new function that fills the hole.

核心要点：Removing a Discontinuity (LIM-2.C.1)

A discontinuity at \(x=c\) is removable if \(\lim\limits_{x \to c} f(x) = L\) exists as a finite number.

To remove the discontinuity, we define a new function \(g(x)\) such that:

\[ g(x) = \begin{cases} f(x), & x \neq c \\ L, & x = c \end{cases} \]

Since the new function value now matches the limit, the function \(g(x)\) is continuous at \(x=c\).

Piecewise Continuity at the Boundary

For piecewise-defined functions, the "boundary" is the \(x\)-value where the formula changes. To ensure continuity across this boundary, we must satisfy the limits from both sides.

核心要点：Boundary Continuity (LIM-2.C.2)

For a piecewise function to be continuous at a boundary \(x=c\):

  1. The limit from the left (\(\lim\limits_{x \to c^-} f(x)\)) must exist.

  2. The limit from the right (\(\lim\limits_{x \to c^+} f(x)\)) must exist.

  3. Both limits must equal each other and equal the function value \(f(c)\).

Example: Filling the Hole

Consider the function \(f(x) = \frac{x^2 - 16}{x - 4}\).

Direct substitution gives \(0/0\) at \(x=4\). Algebraically, we can simplify: $$ f(x) = \frac{(x-4)(x+4)}{x-4} = x+4, \quad x \neq 4 $$

The limit as \(x \to 4\) is \(4+4 = 8\). However, the original function is undefined at \(x=4\). To remove this discontinuity, we redefine the function as:

\[ g(x) = \begin{cases} \frac{x^2 - 16}{x - 4}, & x \neq 4 \\ 8, & x = 4 \end{cases} \]

Extension: Can all discontinuities be removed?

No. Only Removable Discontinuities (holes) can be removed.

  • Jump Discontinuities cannot be removed because the limit does not exist (the two sides don't meet).
  • Infinite Discontinuities cannot be removed because the function values grow to infinity; you cannot "plug" a vertical asymptote with a single point to make it continuous.

Quiz