# What is the limit of x^(1-x) as x approaches 1? The hint is "Use the properties of logarithms to simplify the limit". But I just can think in the straight answer 1, by substitution. Can someone develop in steps using this hint, please?

May 16, 2018

The limit is $1$, as you suspected.

#### Explanation:

It is true you can use substitution but I will show you the logarithm method so that you know how to do it when substitution doesn't work.

$L = {\lim}_{x \to 1} {x}^{1 - x}$

$\ln L = {\lim}_{x \to 1} \ln \left({x}^{1 - x}\right)$

$\ln L = {\lim}_{x \to 1} \left(1 - x\right) \ln x$

$\ln L = 0$

$L = {e}^{0}$

$L = 1$

Hopefully this helps!

May 16, 2018

A slightly different way but with a similar approach

#### Explanation:

As Noah showed, one way to find the limit is to take the natural log of both sides. However, a mistake students might make is that they forget to undo the log. So here's a slightly different way to do it.

${\lim}_{x \to 1} {x}^{1 - x} = {\lim}_{x \to 1} {e}^{\ln} \left({x}^{1 - x}\right)$

$\textcolor{w h i t e}{a} = {e}^{{\lim}_{x \to 1} \left(1 - x\right) \ln x}$

$\textcolor{w h i t e}{a} = {e}^{0}$

$\textcolor{w h i t e}{a} = 1$

This way, we avoid having to undo anything at the end, and we can present our final answer, hopefully, without making any mistakes.