Exercise · ZW-CONTINUOUS-FUNCTION-015 · ★★☆ Intermediate
For \(f(x)=x^2\) at \(c=1\) and \(\varepsilon=0.05\), what is the largest \(\delta\) that works?
Choose one option.
Show answer
- A. \(\delta=\sqrt{1.05}-1\)
Explanation
We need \(0.95<x^2<1.05\), i.e. \(x\in(\sqrt{0.95},\sqrt{1.05})\). The right end is \(\sqrt{1.05}-1\approx0.0247\) away and the left one \(0.0253\); the smaller wins. With \(0.025\) it already leaves the interval.
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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