Answer
See the explanation
Work Step by Step
A function \( f(x) \) can be extended to be continuous at a point \( x = c \) under the following circumstances:
1. The Limit Exists: The limit of \( f(x) \) as \( x \) approaches \( c \) from both the left and the right sides must exist.
2. The Function Value Exists at \( c \): The function \( f(x) \) must have a defined value at \( x = c \).
3. The Limit and Function Value are Equal: The limit of \( f(x) \) as \( x \) approaches \( c \) must be equal to the function value at \( c \).
Mathematically, this can be expressed as:
\[ \lim_{{x \to c^-}} f(x) = \lim_{{x \to c^+}} f(x) \]
\[ \lim_{{x \to c^-}} f(x) = f(c) \]
Example:
Consider the function \( f(x) = \frac{x^2 - 1}{x - 1} \) for \( x \neq 1 \) and \( f(x) = 2 \) for \( x = 1 \). This function is not defined at \( x = 1 \) in its original form due to the division by zero. However, if we simplify the expression by canceling the common factor \( x - 1 \), we get \( f(x) = x + 1 \), which is defined at \( x = 1 \). The limit as \( x \) approaches 1 from both sides is also \( 2 \). Therefore, \( f(x) \) can be extended to be continuous at \( x = 1 \) by defining \( f(1) = 2 \).
