Question

43: Prove that \(\mathop {{\rm{lim}}}\limits_{x \to {0^ + }} {\rm{ln}}x = - \infty \).

Short answer

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

Step-by-step solution

Precise Definition of an infinite limit

Using the definition we have for every \(M < 0\) there exist \(\delta > 0\) such that \(0 < x < \delta \) implies that \({\rm{ln}}x < M\).

Simplify the obtained condition

Rewrite the obtained inequality.

\(\begin{array}{c}{\rm{ln}}x < M\\{e^{{\rm{ln}}x}} < {e^M}\\x < {e^M}\end{array}\)

This implies that \(\delta = {e^M}\).

Therefore, \(0 < x < {e^M}\) and \({\rm{ln}}x < M\) which implies that \({\rm{ln}}x\) tends to \( - \infty \) when \(x\) to \(0\) from the right, that is, \(\mathop {{\rm{lim}}}\limits_{x \to {0^ + }} {\rm{ln}}x = - \infty \).