Question

19-32 Prove the statement using the \(\varepsilon \), \(\delta \)definition of a limit.

25. \(\mathop {{\bf{lim}}}\limits_{x \to 0} {x^{\bf{2}}} = {\bf{0}}\)

Short answer

The given limit is true.

Step-by-step solution

Step 1:Assume the values of \(\varepsilon \) and \(\delta \)

Let \(\varepsilon > 0\) and \(\delta > 0\) such that if \(0 < \left| {x - 0} \right| < \delta \), thenit can be represented as;

\(\left| {{x^2} - 0} \right| < \varepsilon \)

Solve the inequality in step 1

For the inequality\({x^2} < \varepsilon \):

\(\begin{array}{c}{x^2} < \varepsilon \\\left| x \right| < \sqrt \varepsilon \end{array}\)

So, \(\delta = \sqrt \varepsilon \), then \(0 < \left| {x - 0} \right| < \delta \) implies that \(\left| {{x^2} - 0} \right| < \varepsilon \).

Therefore, the limit\(\mathop {\lim }\limits_{x \to a} {x^2} = 0\) is true.