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# 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.$\displaystyle \lim_{x\to \infty} x^{1/x}$

## $$\begin{array}{l}y=x^{1 / x} \Rightarrow \ln y=(1 / x) \ln x \Rightarrow \lim _{x \rightarrow \infty} \ln y=\lim _{x \rightarrow \infty} \frac{\ln x}{x} \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow \infty} \frac{1 / x}{1}=0 \Rightarrow \\\lim _{x \rightarrow \infty} x^{1 / x}=\lim _{x \rightarrow \infty} e^{\ln y}=e^{0}=1\end{array}$$

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we're gonna be taking the limit as X goes to infinity of X to the one over X. Um And we can simplify this by um considering the fact that if we plug in infinity we're going to get infinity to the power of zero. So since that's the indeterminate form, we're going to um take the natural log of both sides. When we do that, we'll end up getting at the natural log of Y equals one over X times the natural log of X, which is just the natural log of acts over X. Then putting infinity, we get infinity over infinity. So we must Take the derivative of both sides giving us one over X Right to Buy one. Now when we plug in infinity we will get one over infinity which is zero divided by one, which is zero. So zero is not our answer. Instead what we have. Is that the natural log of why equals zero is our answer. So let's solve for why we know that we can take E. To the power of both sides. Can I ask why equals E. To the power of zero? So based on this, we see that why is going to equal one? So one ends up being our final final answer. And this was through using a method by which we take the national log on both sides and then employ the hotel's rule.

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