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This question already has an answer here:

It is well known that :$1^{+\infty}$ is indeterminate case ,I have accrossed the following problem which let me to say that :$1^{+\infty}=1$ .

$1^{+\infty}$ can be written as : $1^{+\infty}=\lim _{x\to 0+}(\frac{\sin x}{x})^{1/x}$ which is $ 1$ ,then $1^{+\infty}=1$ and it's not I.case , i don't know where i'm wrong !!!! ? and wolfram alpha says that :$\lim _{x\to 0+}(\frac{\sin x}{x})^{1/x}=1$ which mixed me .

Edit: I have edited the question to show what's mixed me in the side of limit calculation and i don't changed my question

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marked as duplicate by Lord Shark the Unknown, José Carlos Santos, Hans Lundmark, Michael Seifert, Antonio Vargas Jan 14 '18 at 18:42

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    $\begingroup$ If $a_n\to 1$ and $b_n\to\infty$ then $a_n^{b_n}$ may tend to $1$, or may tend to another positive number, or may diverge... $\endgroup$ – Lord Shark the Unknown Jan 14 '18 at 17:54
  • $\begingroup$ "indeterminate" means can not be consistently determined. Determining something in one particular way is not determining it consistently in all possible possible ways. So one interpretation does not contradict indeterminacy. $\endgroup$ – fleablood Jan 14 '18 at 18:38
  • $\begingroup$ But honestly, I'm not seeing why $1^{\infty}$ which I would define as $\lim_{x\to \infty} 1^x$ should be indeterminate at all. So what if $\lim_{y\to 1;x\to \infty} y^x$ is indeterminate. We don't claim $f(a)$ is indeterminate if $f$ is discontinuous at $a$ if it's perfectly well defined. We don't claim $f(a)$ must always equal $\lim_{x\to a} f(x)$ so why should we care about any limits as $y\to 1$? Am I mistaken that $f(\infty)$ should be defined as $\lim_{x\to \infty} f(x)$? $\endgroup$ – fleablood Jan 14 '18 at 19:30
  • $\begingroup$ If you are ok, you can accept the answer and set as solved. Thanks! $\endgroup$ – user Jan 20 '18 at 0:04
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 17:18
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If you think that $$\lim _{x\to 0^+}\bigg(\frac{\sin x}{x}\bigg)^{1/x} = \bigg(\lim_{x \to 0^+}\frac{\sin x}{x}\bigg)^{\lim_{x \to 0^+}1/x}$$ that is not correct. Here, you can find your answer I think: Why is $1^{\infty}$ considered to be an indeterminate form

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What do you think of $$\lim\limits_{x \to \infty} \left(1+\frac{1}{x}\right)^x?$$

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Note that for $a_n \to +\infty$

$$1^{a_n}\to1 \quad (1^{a_n}=1)$$

but for $b_n \to 1$

$$b_n^{a_n}=e^{a_n\log b_n}=e^{+\infty \cdot 0}$$

which is an indeterminate form.

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Indeterminate means we can not assign a single real number to it without causing contradiction

In your question, different examples provide different answers such as $1$ or $e$ , hence the determination is not possible without contradiction.

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  • $\begingroup$ $1^x=1$ for any $x\in R$ so $\lim 1^x=1$ as $x$ tends to $\infty$. But if $lim f(x)^{g(x)}=1^{\infty}$ as $x$ tends to $\infty$($f(x)$ is not constant) is an indeterminate case. $\endgroup$ – M.R. Yegan Jan 14 '18 at 18:34
  • $\begingroup$ Good point for the case of base =1. $\endgroup$ – Mohammad Riazi-Kermani Jan 14 '18 at 18:55

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