I'm working with series and for some reason, it seems that these two limits stop being indeterminate when n is a positive integer. Why is that true ?
Why are $ \lim\limits_{n \to \infty } 1^n = 1$ and $\lim\limits_{n \to \infty } 0^n = 0$ not indeterminate?
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calculus
sequences-and-series
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8Since $1^n=1$ and $0^n=0$, these are *constant* sequences. Why should the limit of a constant sequence be indeterminate? – 2017-01-07
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0Makes sense said this way. I guess I was mistaking constant and approaching a certain value. – 2017-01-07
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2A related, and interesting, question is the form $0^0$. You can find long, pointless and tedious arguments about it. Maybe we would say: $0^0=1$ when the exponent is the integer zero, but $0^0$ is undefined when the exponent is the real number zero. But that would cause even more long, pointless, and tedious arguments. – 2017-01-07
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0Thanks for the help, it's clear now! – 2017-01-07
2 Answers
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Indeterminate forms are indeterminate when both parts are varying. For example, with indeterminate form $$ {1}^{\infty} $$ it is true that when the base does not vary, we have $$ \lim_{x \to +\infty}{1}^{x} = 1 $$ But exponent "positive integer" has nothing to do with it: $$ \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e $$
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When you write it as $\lim\limits_{n\to\infty} 1^n$, you are asking for the limit of the sequence $\{1,1,1,\ldots\}$, which is $1$.
The issue arises when the "$1$" is approaching one but not equal to $1$. So when we talk about the indeterminate form $1^\infty$, that is short hand for the limit of a sequence like $\{(1+1/n)^n\}$. One might be tempted to say that the inside goes to $1$, so sequence eventually looks like $\{1^n\}$. On the other hand, as the exponent gets larger, one might expect the values to blow up. Neither is not true, and the limit actually goes to $e$.