Question

Show that the function

\[ f(x) = \begin{cases} x^2 + 2, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases} \]

is not continuous at \( x = 0 \).

Hint:

A function is continuous at \( x = a \) if \( \lim\limits_{x \to a} f(x) = f(a) \).

Answer 1

Step 1: Check Left-Hand Limit (LHL). \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x^2 + 2) = 2 \] Step 2: Check Right-Hand Limit (RHL). \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 + 2) = 2 \] Step 3: Compare with \( f(0) \). \[ f(0) = 1 \neq 2 \] Since \( LHL \neq f(0) \), the function is not continuous at \( x = 0 \).