Find the limit L. Then use the \(\varepsilon-\delta\) definition to prove that the limit is L.\\
\(\lim_{x \to 0} \sqrt[3]{x}\)
L = 3
Use the \(\varepsilon-\delta\) definition to prove that the limit is L.\
Given \(\varepsilon > 0\), assume \(|f(x) - L| < \delta\), then
\(|\sqrt[3]{x} - \boxed{ }| < \varepsilon\)
\(|x| < \boxed{ }\)
Let \(\delta = \boxed{ }\).
If \(0 < |x| < \delta\) when \(\delta = \varepsilon\)
you have
\(|x| < \varepsilon\)
\(|\sqrt[3]{x} - \boxed{ }| < \varepsilon\)
\(|f(x) - L| < \varepsilon\).
