Exercise · ZW-CONTINUOUS-FUNCTION-020 · ★★★ Advanced
For \(f(x)=1/x\) at \(c=1\), if \(|x-1|<\delta\le1/2\), what is the smallest bound of the form \(k\delta\) valid for \(|f(x)-f(1)|\)?
Choose one option.
Show answer
- C. \(2\delta\)
Explanation
Since \(|1/x-1|=|x-1|/x\) and \(x>1/2\), we get \(<2|x-1|<2\delta\); the ratio approaches \(2\) near \(x=1/2\), so \(2\delta\) is the tightest. Hence \(\delta=\min\{1/2,\varepsilon/2\}\) works.
Review the concept: Continuous function
A function \(f\) is continuous at \(a\) if \(f(a)\) is defined, its limit as \(x\) tends to \(a\) exists and equals \(f(a)\).
\[\lim_{x\to a}f(x)=f(a)\]
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