Question

Find the limit. Use L'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l' Hospital's Rule doesn't apply, explain why.

\(\mathop {\lim }\limits_{x \to \infty } (x - \ln x)\)

Short answer

The value of \(\mathop {\lim }\limits_{x \to - \infty } (x - \ln x)\) is \(\infty \).

Step-by-step solution

Given information

The given function is \(\mathop {\lim }\limits_{x \to - \infty } (x - \ln x)\).

Concept of L' Hospital's Rule

Suppose that one of the following cases.

\(\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{0}{0}\quad {\rm{ OR }}\quad \mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{ \pm \infty }}{{ \pm \infty }}\)

Here\(a\)can be any real number, infinity, or negative infinity.

In these cases,\(\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \frac{{{f^\prime }(x)}}{{{g^\prime }(x)}}\).

Indeterminate form of given function

Express given function as shown below.

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

Therefore, \(\mathop {\lim }\limits_{x \to - \infty } (x - \ln x) = \mathop {\lim }\limits_{x \to \infty } x \cdot \mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{{\ln x}}{x}} \right)\). …… (1)

Consider, \(\mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{{\ln x}}{x}} \right) = \mathop {\lim }\limits_{x \to \infty } 1 - \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{x}\).

Obtain the value of the function as \(x\) approaches \(\infty \).

The limit of the first term is, \(\mathop {\lim }\limits_{x \to \infty } 1 = 1\).

As \(x\) approaches \(\infty \), the numerator of the second term \(\ln x\) approaches to \(\infty \) and denominator of the second term \(x\) also approaches to \(\infty \).

Thus, \(\mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{{\ln x}}{x}} \right) = \frac{\infty }{\infty }\) is in an indeterminate form.

Apply L' Hospital's Rule

Therefore, apply L'Hospital's Rule and obtain the limit as shown below.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{{\ln x}}{x}} \right) = \mathop {\lim }\limits_{x \to \infty } 1 - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\frac{1}{x}}}{1}} \right)\\ = 1 - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right)\\ = 1 - \frac{1}{\infty }\\ = 1\end{array}\)

Thus, \(\mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{{\ln x}}{x}} \right) = 1\). ……. (2)

Substitute the equation (2) in the equation (1).

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

Thus, the limit of the function is \(\infty \).