Question

Prove that \(\mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{{{x^p}}} = 0\) for any number \(p > 0\). This shows that the logarithmic function approaches \(\infty \) more slowly than any power of \(x\).

Short answer

The limit is of the form \(\frac{\infty }{\infty }\), it is proved that \(\mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{{{x^p}}} = 0\).

Step-by-step solution

Given information

The Logarithmic function approaches \(\infty \) more slowly than any power of \(x\).

Concept of L Hopital’s Rule

If two functions in the neighborhood of a given point have an infinite limit or zero as a limit and are both differentiable.

If this limit exists, the limit of the quotient of the functions is equal to the limit of the quotient of their derivatives.

Apply L Hospital’s Rule for solution

Consider the logarithmic function approaches \(\infty \) more slowly than any power of \(x\).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \ln x = \infty \\\mathop {\lim }\limits_{x \to \infty } {x^p} = \infty ,p > 0\end{array}\)

Since the limit is of the form \(\frac{\infty }{\infty }\), apply L’Hospital's Rule.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{{{x^p}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{x}}}{{p{x^{p - 1}}}}\\ = \frac{1}{p}\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} \cdot \frac{1}{{{x^{p - 1}}}}\\ = \frac{1}{p}\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^{p - 1 + 1}}}}\\ = \frac{1}{p}\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^p}}}\end{array}\)

Apply the limit as shown below.

\(\begin{array}{c} = \frac{1}{\infty }\\ = 0\end{array}\)

Hence, proved.

Since the limit is of the form \(\frac{\infty }{\infty }\), apply L’Hospital’s Rule once.