Chapter 8.2, Problem 47E

To determine

To calculate: The limit of the given expression $\underset{x\to 1^{+}}{\lim }{x}^{1/\left(1-x\right)}$ .

Answer to Problem 47E

The limit of the function is 1.

Explanation of Solution

Given Information:

The expression is defined as,

  $\underset{x\to 1^{+}}{\lim }{x}^{1/\left(1-x\right)}$

Formula Used:

L’hospital rule is defined as,

If the limit is giving undefined form $\frac{\infty }{\infty },\frac{0}{0}$ for the function $\frac{f\left(x\right)}{g\left(x\right)}$ .

  $\frac{f\left(x\right)}{g\left(x\right)}=\frac{\frac{d}{dx}\left(f\left(x\right)\right)}{\frac{d}{dx}\left(g\left(x\right)\right)}$

Calculation:

Consider the given fraction,

  $\underset{x\to 1^{+}}{\lim }{x}^{1/\left(1-x\right)}$

First find the limit of the expression.

  $\begin{array}{c}\underset{x\to 1^{+}}{\lim }{x}^{1/\left(1-x\right)}=\underset{x\to 0^{+}}{\lim }{e}^{\ln x^{1/\left(1-x\right)}}\\ =\underset{x\to 0^{+}}{\lim }{e}^{\frac{1}{1-x}\ln x}\\ =e^{\underset{x\to 0^{+}}{\lim }\frac{1}{1-x}\ln x}\end{array}$

Further simplify the limit and use L’hospital rule.

  $\begin{array}{c}\underset{x\to 1^{+}}{\lim }\frac{1}{1-x}\ln x=\underset{x\to 1^{+}}{\lim }\frac{\ln x}{1-x}\\ =\underset{x\to 1^{+}}{\lim }\frac{\frac{1}{x}}{-1}\left[\text{L'hospital rule}\right]\\ =\underset{x\to 1^{+}}{\lim }\left(-\sin x\cos x\right)\\ =0\end{array}$

Resubstitute the value of the obtained limit.

  $\begin{array}{c}\underset{x\to 0^{+}}{\lim }{\left(\sin x\right)}^{\tan x}=e^0\\ =1\end{array}$

Hence, the limit of the function is 1.