Question

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

28. \(\mathop {\lim }\limits_{x \to - {6^ + }} \sqrt(8){{6 + x}} = 0\)

Short answer

It is proved that \(\mathop {\lim }\limits_{x \to - {6^ + }} \sqrt(8){{6 + x}} = 0\).

Step-by-step solution

Precise Definition of Left-hand and Right-hand limit

Consider \(\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = L\).

If for all numbers \(\varepsilon > 0\) there exists a number \(\delta > 0\) such that if \(a - \delta < x < a\) then \(\left| {f\left( x \right) - L} \right| < \varepsilon \).

Consider\(\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = L\).

If for all numbers \(\varepsilon > 0\) there exists a number \(\delta > 0\) such that if \(a < x < a + \delta \) then \(\left| {f\left( x \right) - L} \right| < \varepsilon \).

Prove the statement using the definition of a limit

Let \(\varepsilon > 0\)be a given positive number. We want \(\delta > 0\) such that when\(0 < x - \left( { - 6} \right) < \delta \) then \(\left| {\sqrt(8){{6 + x}} - 0} \right| < \varepsilon \). However,

\(\begin{array}{c}\left| {\sqrt(8){{6 + x}} - 0} \right| < \varepsilon \\\sqrt(8){{6 + x}} < \varepsilon \\6 + x < {\varepsilon ^8}\\x - \left( { - 6} \right) < {\varepsilon ^8}\end{array}\)

Therefore, when we choose \(\delta = {\varepsilon ^8}\) then\(0 < x - \left( { - 6} \right) < \delta \Rightarrow \left| {\sqrt(8){{6 + x}} - 0} \right| < \varepsilon \)

Hence, \(\mathop {\lim }\limits_{x \to - {6^ + }} \sqrt(8){{6 + x}} = 0\) according to the definition of a right-hand limit.

Thus, it is proved that \(\mathop {\lim }\limits_{x \to - {6^ + }} \sqrt(8){{6 + x}} = 0\).