# `lim_(x->1^+) x^(1/(1-x))` 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,...

`lim_(x->1^+) x^(1/(1-x))` 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.

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You need to evaluate the limit, using logarithm special technique, such that:

`f(x) = x^(1/(1-x))ln f(x) = ln (x^(1/(1-x)))`

`ln f(x) = 1/(1-x)*ln x`

`lim_(x->1^+) 1/(1-x)*ln x = 0/0`

Since the limit is indeterminate `0/0` , you may use l'Hospital's rule:

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

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

**Hence, evaluating the limit using l'Hospital's rule, yields `lim_(x->1^+) x^(1/(1-x)) = 1/e.` **

`x^(1/(1-x))=e^(1/(1-x)*lnx)`

`=lim_(x->1^+)(e^(1/(1-x)*lnx))`

Use the chain rule for limits

let` g(x)=(1/(1-x))lnx and f(u)=e^u`

`lim_(x-gt1^+)(1/(1-x)*lnx)=-1`

`lim_(u-gt-1)(e^u)=1/e`

Thus, the answer is `1/e`