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 \( x = 0 \).
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 \( x = 0 \).Show Hint
A function is continuous at \( x = a \) if \( \lim\limits_{x \to a} f(x) = f(a) \).
Updated On: Feb 27, 2025
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Solution and Explanation
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 \( x = 0 \).
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