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.

37. \(\mathop {\lim }\limits_{x \to {1^ + }} {x^{1/(1 - x)}}\)

Short answer

The original limit is \(\mathop {\lim }\limits_{x \to {1^ + }} {x^{1/(1 - x)}}\)is\(\frac{1}{e}\).

Step-by-step solution

Given information

The function is \(\mathop {\lim }\limits_{x \to {1^ + }} {x^{1/(1 - x)}}\).

Concept of L Hospital’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.

Find the limit

Consider,\(y = \mathop {\lim }\limits_{x \to {1^ + }} {x^{\frac{1}{{1 - x}}}}\).

Take \(In\) on both sides,

\(\begin{aligned}\ln y &= \mathop {\lim }\limits_{x \to {1^ + }} \ln {x^{1/(1 - x)}}\\ &= \mathop {\lim }\limits_{x \to {1^ + }} \frac{1}{{1 - x}}\ln x\end{aligned}\)

Put\(x = 1\).

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

Since \(1 - x\) is nonzero near 1, and so is its' derivative \( - 1\), use L'Hospital's rule.

Apply L’Hopital’s rule and solve

Differentiate the numerator and the denominator.

\(\begin{aligned}\ln y &= \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\frac{d}{{dx}}\ln x}}{{\frac{d}{{dx}}(1 - x)}}\\ &= \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\frac{1}{x}}}{{ - 1}}\\ &= - \frac{1}{1}\\ &= - 1\end{aligned}\)

Therefore, \(\ln y = - 1\).

To find\(y\), take the exponent of both sides.

\(\begin{array}{c}y = {e^{ - 1}}\\ = \frac{1}{e}\end{array}\)