Thread: Limit of Indeterminate form 1^(infinity)

1. Limit of Indeterminate form 1^(infinity)

A question on my homework is: lim(x->1+) (x^(1/(1-x)) The given answer is 0.367879 and by inspection of a graphing calculator given graph I can see that this is indeed correct, however when I attempt to find the limit: I set it up as a logarithm: ln y = ln(x^(1/(1-x)) = (1/(1-x)) ln(x) = ln(x) / (1-x) Then I use L'Hopital's Rule (since it is now of the indeterminate form 0/0): (1/x) / 1 which yields 1. We must remember to undo turning it into a logarithm so: e^1 = e but e is not the answer. >:/ How am I screwing this up?

2. Hello
Originally Posted by InfinitePartsInHarmony
A question on my homework is: lim(x->1+) (x^(1/(1-x)) The given answer is 0.367879 and by inspection of a graphing calculator given graph I can see that this is indeed correct, however when I attempt to find the limit: I set it up as a logarithm: ln y = ln(x^(1/(1-x)) = (1/(1-x)) ln(x) = ln(x) / (1-x) Then I use L'Hopital's Rule (since it is now of the indeterminate form 0/0): (1/x) / 1 which yields 1. We must remember to undo turning it into a logarithm so: e^1 = e but e is not the answer. >:/ How am I screwing this up?
Here is your mistake : $\displaystyle (1-x)'={\color{red}-}1$. Using this, you'll find that the answer is $\displaystyle \mathrm{e}^{-1}=\frac{1}{\mathrm{e}}$.

3. Originally Posted by InfinitePartsInHarmony
A question on my homework is: lim(x->1+) (x^(1/(1-x)) The given answer is 0.367879 and by inspection of a graphing calculator given graph I can see that this is indeed correct, however when I attempt to find the limit: I set it up as a logarithm: ln y = ln(x^(1/(1-x)) = (1/(1-x)) ln(x) = ln(x) / (1-x) Then I use L'Hopital's Rule (since it is now of the indeterminate form 0/0): (1/x) / 1 which yields 1. We must remember to undo turning it into a logarithm so: e^1 = e but e is not the answer. >:/ How am I screwing this up?
You have made a slight typo:

$\displaystyle \frac{\ln(x)}{1-x}~\underbrace{\implies}_{L'Hopital's\ rule}~{\color{red} - }\frac{\frac1x}{1}$ which yields -1 if x approaches 1.

Therefore the limit is $\displaystyle e^{-1} = \frac1e$

1^infinity indeterminate form

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