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It isn’t: $\lim_n\to\infty1^n=1$, specifically as friend suggest. However, if $f$ and also $g$ are features such that $\lim_n\to\inftyf(n)=1$ and also $\lim_n\to\inftyg(n)=\infty$, the is

**not**necessarily true that

$$\lim_n\to\inftyf(n)^g(n)=1\;.\tag1$$

For example, $$\lim_n\to\infty\left(1+\frac1n\right)^n=e\approx2.718281828459045\;.$$

More generally,

$$\lim_n\to\infty\left(1+\frac1n\right)^an=e^a\;,$$

and together $a$ varieties over all real numbers, $e^a$ ranges over all positive real numbers. Finally,

$$\lim_n\to\infty\left(1+\frac1n\right)^n^2=\infty\;,$$

and

$$\lim_n\to\infty\left(1+\frac1n\right)^\sqrt n=0\;,$$

so a border of the form $(1)$ always has come be evaluated on its own merits; the borders of $f$ and also $g$ don’t by themselves recognize its value.

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answer Mar 3 "13 in ~ 20:29

Brian M. ScottBrian M. Scott

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The limit of $1^\infty$ exist:$$\lim_n\to\infty1^n$$ is no indeterminate. However$$\lim_a\to 1^+,n\to\inftya^n$$ is indeterminate..

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edited Apr 15 "18 in ~ 4:43

KingLogic

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answered Mar 3 "13 in ~ 20:21

user35039user35039

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There are many reasons. Because that example, permit $1^\infty=1$. Taking logarithm, you have actually $\infty\cdot 0=0$. Likewise for other operations friend will achieve some absurd.

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reply Mar 3 "13 in ~ 20:25

Boris NovikovBoris Novikov

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The brief answer is the it is due to the fact that $x^y$ can tend to any type of nonnegative limit as $x\to1$ and $y\to\infty$. Because that one example, consider the timeless limit $$\lim_n\to\infty\Bigl(1+\frac xn\Bigr)^n=e^x.$$

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edited Apr 13 "18 at 0:27

Paul Sinclair

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answer Mar 3 "13 in ~ 20:20

Harald Hanche-OlsenHarald Hanche-Olsen

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