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Numerade Educator

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Problem 61 Hard Difficulty

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.

$ \displaystyle \lim_{x\to 1^+} x^{1/(1 - x)} $

Answer

$$\begin{array}{l}
y=x^{1 /(1-x)} \Rightarrow \ln y=\frac{1}{1-x} \ln x, \text { so } \lim _{x \rightarrow 1^{+}} \ln y=\lim _{x \rightarrow 1^{+}} \frac{1}{1-x} \ln x=\lim _{x \rightarrow 1^{-}} \frac{\ln x}{1-x} \underline{\Perp}=\lim _{x \rightarrow 1^{+}} \frac{1 / x}{-1}=-1 \Rightarrow \\
\lim _{x \rightarrow 1^{+}} x^{1 /(1-x)}=\lim _{x \rightarrow 1^{+}} e^{\ln y}=e^{-1}=\frac{1}{e}
\end{array}$$

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Video Transcript

because people were taking the limit as X goes to one from the right of the function X to the power of 1/1 -1. So that is the function that we're dealing with here. Um And We see that when we plug in one we get uh well first we have to take the natural log of both sides. When we do that, that will allow us to get it in the form 1/1 -1 times natural log effects. And when we evaluate that at one we see we get the indeterminant form. So we need to use like tiles room. This will allow us to have um If we write it up here natural log of X, it will be one over X. This will be just a -1. Okay then we plug in the one, we get one a negative one. So our answer is going to be negative one. Remember that? That's the natural log. So our natural log of Y is going to equal negative one. Therefore, when we solve for Y, we'll end up getting that Y equals E. to the -1, which is the same thing as one over E. So one over E is going to be our final answer.