Question

Find the limit. Use l’Hospital’s Rule where app\( - 1\)ropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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

Short answer

The value of the limit is .

Step-by-step solution

Definition of l’Hospital’s Rule

Let \(f\) and \(g\) be differentiable and \(g'\left( x \right) \ne 0\) on an open interval I, containing \(a\) (except possibly at \(a\)). Assume that \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = 0\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = 0\), or \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = \pm \infty \), and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = \pm \infty \).

(In certain, there is anindeterminate formof type \(\frac{0}{0}\) or \(\frac{\infty }{\infty }\).) Then;

\(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\)

The limit exists on the right-hand side (or is \(\infty \) or \( - \infty \)).

Determine the limit

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} {x^2} = 0,{\rm{ }}\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{x} = \infty \,\,\,\,{\mathop{\rm and}\nolimits} \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} {x^2} \cdot \frac{1}{x} = \mathop {\lim }\limits_{x \to {0^ + }} {x^2} = 0\\\mathop {\lim }\limits_{x \to {0^ + }} x = 0,{\rm{ }}\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{{x^2}}} = \infty \,\,\,{\mathop{\rm and}\nolimits} \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} x \cdot \frac{1}{{{x^2}}} = \infty \end{array}\)

The above limits are known as an indeterminate form of type\(0 \cdot \infty \). Solve the limit by expressing the product \(fg\) as a quotient as shown below:

\(fg = \frac{f}{{\frac{1}{g}}}\) or \(fg = \frac{g}{{\frac{1}{f}}}\)

This changes the provided limit into an indeterminate form of type \(\frac{0}{0}\) or \(\frac{\infty }{\infty }\).

Evaluate the limit as shown below:

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

It is observed that the limit is in indeterminate product form \( - \infty \cdot 0\).

Write \(x = \frac{1}{{\frac{1}{x}}}\) in the limit as shown below:

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

It is observed that the limit is in indeterminate form \(\frac{0}{0}\).

Apply l’Hospital Rule as shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to - \infty } \frac{{\ln \left( {1 - \frac{1}{x}} \right)}}{{\frac{1}{x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left( {\ln \left( {1 - \frac{1}{x}} \right)} \right)}}{{\frac{d}{{dx}}\left( {\frac{1}{x}} \right)}}\\ = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{{1 - \frac{1}{x}}} \cdot \frac{1}{{{x^2}}}}}{{ - \frac{1}{{{x^2}}}}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{1}{{1 - \frac{1}{x}}}}}{{ - 1}}\\ = \mathop {\lim }\limits_{x \to - \infty } - \frac{1}{{1 - \frac{1}{x}}}\\ = - \frac{1}{1}\\ = - 1\end{array}\)

Thus, the value of the limit is \( - 1\).