Tutorial Video:Continuity -Example 5
Engineering Mathematica: R-2025 Updates
Example: Show that the function \(f(x)\) is continuous on \((-\infty,\infty)\):
\[ f(x) = \begin{cases} 1-x^2, & x \leq 1 \\ \log x, & x > 1 \end{cases} \]
[Reference: N/D-22-R-21-CUR]
Detailed Solution
To prove continuity at \(x = a\), we must satisfy the condition:
\[ \lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) = f(a) \]
Checking continuity at the critical point \(x = 1\):
Left-Hand Limit (LHL):
\[ \lim_{x \to 1^{-}} (1-x^2) = 1-(1)^2 = 0 \] Right-Hand Limit (RHL):
\[ \lim_{x \to 1^{+}} (\log x) = \log(1) = 0 \] Function Value:
\[ f(1) = 1-(1)^2 = 0 \]
\[ \lim_{x \to 1^{-}} (1-x^2) = 1-(1)^2 = 0 \] Right-Hand Limit (RHL):
\[ \lim_{x \to 1^{+}} (\log x) = \log(1) = 0 \] Function Value:
\[ f(1) = 1-(1)^2 = 0 \]
Since the LHL = RHL = \(f(1) = 0\), the function is continuous at \(x = 1\).
Because the individual functions (\(1-x^2\) and \(\log x\)) are continuous on their respective domains, the entire function \(f(x)\) is continuous on \((-\infty, \infty)\).
\[ \therefore f(x) \text{ is continuous on } (-\infty, \infty). \]
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