Approaching in the conventional way, as n approaches infinity, n/(n+1) approaches 1 and 1^infinity is 1. so the limit is 1.
But usually, I have seen it solved as 1/e which makes sense as well.
Why isn't the first approach incorrect if it is?
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1Welcome to MSE. Please, use MathJax to format your question: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Green.H Jun 03 '18 at 13:08
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In the language of calculus textbooks, $1^\infty$ is an indeterminate form. Look for that in the index of your textbook. – GEdgar Jun 03 '18 at 13:32
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The first approach doesn’t work because $\frac{n}{n+1}$ is actually a little bit smaller than $1$. It gets closer and closer to $1$ as $n\to\infty$, but the exponent gets bigger and bigger, too. So you must determine which one approaches faster... if the exponent grows too fast, the limit will be $0$. If the fraction converges to $1$ more rapidly, the limit can be $1$. Or their rates could “compromise” and give you something between $0$ and $1$ (in this case, you do indeed get $1/e$.
Clayton
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