Question

Prove that the function \[ f(x) = \begin{cases} x^3 - 3 & \text{if } x \leq 2 \\ x^2 + 1 & \text{if } x>2 \end{cases} \] is continuous at \( x = 2 \).

Hint:

To prove continuity, verify the left-hand and right-hand limits as well as the value of the function at the point.

Answer 1

Step 1: Definition of Continuity.
A function is continuous at a point \( x = c \) if the following condition holds: \[ \lim_{x \to c} f(x) = f(c). \] In this case, we need to prove that the function is continuous at \( x = 2 \), meaning we must show: \[ \lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x). \]
Step 2: Left-hand Limit.
For \( x \leq 2 \), we use the expression \( f(x) = x^3 - 3 \). The left-hand limit is: \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^3 - 3) = 2^3 - 3 = 8 - 3 = 5. \]
Step 3: Right-hand Limit.
For \( x>2 \), we use the expression \( f(x) = x^2 + 1 \). The right-hand limit is: \[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 + 1) = 2^2 + 1 = 4 + 1 = 5. \]
Step 4: Value of the Function at \( x = 2 \).
Since \( x = 2 \) is in the domain where \( x \leq 2 \), we use the expression \( f(x) = x^3 - 3 \) to find: \[ f(2) = 2^3 - 3 = 8 - 3 = 5. \]
Step 5: Conclusion.
We have shown that: \[ \lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x) = 5. \] Thus, the function is continuous at \( x = 2 \).