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.

61. \(\mathop {\lim }\limits_{x \to {1^ + }} {x^{{1 \mathord{\left/

{\vphantom {1 {\left( {1 - x} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {1 - x} \right)}}}}\)

Short answer

The value of \(\mathop {\lim }\limits_{x \to {1^ + }} {\left( x \right)^{{1 \mathord{\left/

{\vphantom {1 {\left( {1 - x} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {1 - x} \right)}}}}\) is \(\frac{1}{e}\).

Step-by-step solution

L’Hospital’s Rule

Suppose there are two differentiable functions \(f\) and \(g\) on any open interval \(I\) which contains \(a\). Then 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 \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = \pm \infty \), or it can be said that if there is an indeterminate form of kind \(\frac{0}{0}\) and \(\frac{\infty }{\infty }\), then \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\).

Indeterminate forms

Type 1: \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{0}{0}\)

Type 2: \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{\infty }{\infty }\)

Type 3: Indeterminate Products, \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) \cdot g\left( x \right)} \right) = 0 \cdot \infty \)

Type 4: Indeterminate difference, \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - g\left( x \right)} \right) = \infty - \infty \)

Type 5: Indeterminate Powers, \(\mathop {\lim }\limits_{x \to a} {\left( {f\left( x \right)} \right)^{g\left( x \right)}} = {0^0},{\infty ^0},{1^\infty }\)

Apply limits directly

The given limit function is \(\mathop {\lim }\limits_{x \to {1^ + }} {\left( x \right)^{{1 \mathord{\left/

{\vphantom {1 {\left( {1 - x} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {1 - x} \right)}}}}\).

Solve \(\mathop {\lim }\limits_{x \to {1^ + }} {\left( x \right)^{{1 \mathord{\left/

{\vphantom {1 {\left( {1 - x} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {1 - x} \right)}}}}\).

\(\mathop {\lim }\limits_{x \to {1^ + }} {\left( x \right)^{{1 \mathord{\left/

{\vphantom {1 {\left( {1 - x} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {1 - x} \right)}}}} = {1^\infty }\)

The above limit value is an indeterminate form, for which L’Hospital’s Rule cannot be applied to solve further.

Simplify the limit function

Let \(y = {x^{{1 \mathord{\left/

{\vphantom {1 {\left( {1 - x} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {1 - x} \right)}}}}\).

Apply a natural log on both sides.

\(\begin{array}{c}\ln y = \ln {x^{{1 \mathord{\left/

{\vphantom {1 {\left( {1 - x} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {1 - x} \right)}}}}\\ = \frac{1}{{\left( {1 - x} \right)}}\ln x\end{array}\)

Then apply a limit on both sides.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {1^ + }} \left( {\ln y} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{{\ln x}}{{\left( {1 - x} \right)}}} \right)\\ = \frac{0}{0}\end{array}\)

The above form is an indeterminate form of type \(\frac{0}{0}\).

Apply L’Hospital Rule to solve

Apply L’Hospital’s Ruleon\(\mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{{\ln x}}{{\left( {1 - x} \right)}}} \right)\)and simplify.

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

Now, solve for the given limits.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {1^ + }} {\left( x \right)^{{1 \mathord{\left/

{\vphantom {1 {\left( {1 - x} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {1 - x} \right)}}}} = \mathop {\lim }\limits_{x \to {1^ + }} \left( {{e^{\ln y}}} \right)\\ = {e^{ - 1}}\\ = \frac{1}{e}\end{array}\)

So, the value of\(\mathop {\lim }\limits_{x \to {1^ + }} {\left( x \right)^{{1 \mathord{\left/

{\vphantom {1 {\left( {1 - x} \right)}}} \right.

\kern-\nulldelimiterspace} {\left( {1 - x} \right)}}}}\) is \(\frac{1}{e}\).