Question

44: (a) For \(f\left( x \right) = \frac{2}{x} - \frac{1}{{\ln x}}\) find each of the following limits.

1. \(\mathop {\lim }\limits_{x \to \infty } f\left( x \right)\)
2. \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\)
3. \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)\)
4. \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\)

(b) Use the information from part (a) to make a rough sketch of the graph of \(f\).

Short answer

(a)

(i) The solution of the limit is \(\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 0\).

(ii)The solution of the limit is \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \infty \).

(iii)The solution of the limit is \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \infty \).

(iv)The solution of the limit is \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = - \infty \).

(b) The graph of the function is shown below:

Step-by-step solution

Evaluate the limits

(a)

(i)

Evaluate the limit as shown below:

\(\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{2}{x} - \frac{1}{{\ln x}}} \right)\)

Since \(\,\frac{2}{x} \to 0\), and \(\frac{1}{{\ln x}} \to 0\) as \(x \to \infty \). So, the value of the limit is shown below:

\(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{2}{x} - \frac{1}{{\ln x}}} \right) = 0\)

Thus, the solution of the limit is \(\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 0\).

(ii)

Evaluate the limit as shown below:

\(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{2}{x} - \frac{1}{{\ln x}}} \right)\)

Since \(\,\frac{2}{x} \to \infty \), and \(\frac{1}{{\ln x}} \to 0\) as \(x \to {0^ + }\). So, the value of the limit is shown below:

\(\mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{2}{x} - \frac{1}{{\ln x}}} \right) = \infty \)

Thus, the solution of the limit is \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \infty \).

(iii)

Evaluate the limits as shown below:

\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {\frac{2}{x} - \frac{1}{{\ln x}}} \right)\)

Since \(\,\frac{2}{x} \to 2\), and \(\frac{1}{{\ln x}} \to - \infty \) as \(x \to {1^ - }\). So, the value of the limit is shown below:

\(\mathop {\lim }\limits_{x \to {1^ - }} \left( {\frac{2}{x} - \frac{1}{{\ln x}}} \right) = \infty \)

Thus, the solution of the limit is \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \infty \).

(iv)

Evaluate the limits as shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{2}{x} - \frac{1}{{\ln x}}} \right)\left( {{\mathop{\rm since}\nolimits} \,\,\frac{2}{x} \to 2\,\,{\mathop{\rm and}\nolimits} \,\,\frac{1}{{\ln x}} \to \infty \,\,{\mathop{\rm as}\nolimits} \,\,x \to {1^ + }} \right)\\ = - \infty \end{array}\)

Since \(\,\frac{2}{x} \to 2\), and \(\frac{1}{{\ln x}} \to \infty \) as \(x \to {1^ + }\). So, the value of the limit is shown below:

\(\mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{2}{x} - \frac{1}{{\ln x}}} \right) = - \infty \)

Thus, the solution of the limit is \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = - \infty \).

Use the information from part (a) to sketch the graph

It is observed from part (a) that \(x = 1\).

Sketch the graph of the function \(f\left( x \right) = \frac{2}{x} - \frac{1}{{\ln x}}\) by using the information from part (a) and the line \(x = 1\) as shown below:

Hence, the graph of the function is obtained.