Question

Test whether the function $f : \mathbb{R} \to \mathbb{R}$ defined by \[f(x) = \begin{cases} x^3 + 3, & x \neq 0 \\ 1, & x = 0 \end{cases}\] is continuous at $x=0$.

Hint:

For continuity at $x=a$: $\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = f(a)$.

Answer 1

Step 1: Value of the function at $x=0$.
\[ f(0) = 1 \]

Step 2: Left-hand and right-hand limits as $x \to 0$.
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x^3+3) = 3 \] \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^3+3) = 3 \] So, \[ \lim_{x \to 0} f(x) = 3 \]

Step 3: Compare limit with $f(0)$.
\[ \lim_{x \to 0} f(x) = 3 \neq f(0) = 1 \]

Step 4: Conclusion.
Since $\lim_{x \to 0} f(x) \neq f(0)$, \[ \boxed{f(x) \;\text{is not continuous at}\; x=0} \]