Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 20

Show that the statement $\lim \limits_{x \to 0} \sqrt[3]{x} = 0$ using the precise definition of a limit.

Based from the definition,

$\qquad$ if $0 < | x - a | < \delta $ then $|f(x) - L | < \varepsilon $

$\qquad$ if $0 < | x -0 | < \delta $ then $|\sqrt[3]{x} - 0| < \varepsilon$

That is,

$\qquad$ if $0 < | x | < \delta$ then $|\sqrt[3]{x}| < \varepsilon$

Or taking the cube of both sides of the inequality $|\sqrt[3]{x}| < \varepsilon$, we get

$\qquad$ if $ 0 < |x| < \delta$ then $|x| < \varepsilon^3$

The statement suggest that we should choose $\displaystyle \delta = \varepsilon^3$.

By proving that the assumed value of $\displaystyle \delta = \varepsilon^3$ will fit the definition.

$\qquad$ if $0 < |x| < \delta$ then,

$\qquad \quad$ $|\sqrt[3]{x}| < \sqrt[3]{\delta} = \sqrt[3]{\varepsilon^3} = \varepsilon$

Thus,

$\qquad$ if $0 < | x - 0 | < \delta $ then $|\sqrt[3]{x} -0 | < \varepsilon$

Therefore, by the precise definition of a limit

$\qquad$ $\lim \limits_{x \to 0} \sqrt[3]{x} = 0$