Question

Use Definition 8 to prove that \(\mathop {{\rm{lim}}}\limits_{x \to - \infty } \frac{1}{x} = 0\).

Short answer

It is proved that \(\mathop {{\rm{lim}}}\limits_{x \to - \infty } \frac{1}{x} = 0\).

Step-by-step solution

Precise Definition of a limit at infinity  

Assume \(f\) be any function defined on the interval \(\left( { - \infty ,a} \right)\). Then \(\mathop {{\rm{lim}}}\limits_{x \to - \infty } f\left( x \right) = L\), which implies that for every \( \in > 0\) there exist a number \(N\) such that if \(x < N\) then \(\left| {f\left( x \right) - L} \right| < \in \).

Given \( \in > 0\), we have to find \(N\) such that if \(x < N\) then \(\left| {\frac{1}{x} - 0} \right| < \in \).

For \(x < 0\), \(\left| {\frac{1}{x} - 0} \right| = - \frac{1}{x} < \in \).

Simplify the inequality

Simplify the inequality \( - \frac{1}{x} < \in \) for \(x\).

\(\begin{array}{c} - \frac{1}{x} < \in \\x < - \frac{1}{ \in }\end{array}\)

Choose \(N = - \frac{1}{ \in }\) which satisfy if \(x < - \frac{1}{ \in }\) then \(\left| {\frac{1}{x} - 0} \right| = - \frac{1}{x} < \in \).

By using Precise Definition of a limit at infinity, \(\mathop {{\rm{lim}}}\limits_{x \to - \infty } \frac{1}{x} = 0\).

Hence, it is proved that \(\mathop {{\rm{lim}}}\limits_{x \to - \infty } \frac{1}{x} = 0\).