Question

Prove the statement using the\(\varepsilon \), \(\delta \)definition of a limit.

\(\mathop {\lim }\limits_{x \to 0} {x^3} = 0\)

Short answer

It is proved that\(\mathop {\lim }\limits_{x \to 0} {x^3} = 0\).

Step-by-step solution

Describe the given information

It is required to prove\(\mathop {\lim }\limits_{x \to 0} {x^3} = 0\)by using\(\varepsilon \),\(\delta \)definition.

Prove that \(\mathop {\lim }\limits_{x \to 0} {x^3} = 0\)

Consider the limit\(\mathop {\lim }\limits_{x \to 0} {x^3} = 0\).

According to the definition with \(a = 0\)and\(L = 0\), it is required to find a number \(\delta \) such that if \(0 < \left| {x - 0} \right| < \delta \) then\(\left| {{x^3} - 0} \right| < \varepsilon \).

If \(0 < \left| x \right| < \delta \)then\(\left| {{x^3}} \right| < \varepsilon \).

If \(\delta = {\left( \varepsilon \right)^{\frac{1}{3}}}\) is chosen then it happens that if \(0 < \left| x \right| < \delta \)then\(\left| {{x^3}} \right| < {\left( {{{\left( \varepsilon \right)}^{\frac{1}{3}}}} \right)^3} = \varepsilon \). Therefore, by the definition of limit\(\mathop {\lim }\limits_{x \to 0} {x^3} = 0\).

Therefore, it is proved that\(\mathop {\lim }\limits_{x \to 0} {x^3} = 0\).