What is the lim_(xrarr1^+) x^(1/(1-x)) as x approaches 1 from the right side?

1 Answer
Jun 14, 2017

1/e


x^(1/(1-x)):

graph{x^(1/(1-x)) [-2.064, 4.095, -1.338, 1.74]}

Well, this would be much easier if we simply took the ln of both sides. Since x^(1/(1-x)) is continuous in the open interval to the right of 1, we can say that:

ln [lim_(x->1^(+)) x^(1/(1-x))]

= lim_(x->1^(+)) ln (x^(1/(1-x)))

= lim_(x->1^(+)) ln x/(1-x)

Since ln(1) = 0 and (1 - 1) = 0, this is of the form 0/0 and L'Hopital's rule applies:

= lim_(x->1^(+)) (1"/"x)/(-1)

And of course, 1/x is continuous from each side of x = 1.

=> ln [lim_(x->1^(+)) x^(1/(1-x))] = -1

As a result, the original limit is:

color(blue)(lim_(x->1^(+)) x^(1/(1-x))) = "exp"(ln [lim_(x->1^(+)) x^(1/(1-x))])

= e^(-1)

= color(blue)(1/e)