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

1 Answer
Truong-Son N.
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)#

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