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How to evaluate the limit: $$\lim_{x \to 0} \Bigl(\frac{\sin{x}}{x}\Bigr)^{1/x^{3}}$$

I think it goes to $1$ because $\lim\limits_{x \to 0} \frac{\sin{x}}{x} =1$ and so power of $1$ should also be $1$. Am I right?

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    No $1^{\infty}$ is considered to be an indeterminate form.2011-09-03
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    For comparison, consider the standard limit $e = \lim_{x \to 0} (1+x)^{1/x}$. Applying your logic would give $1$ as the limit. (By the way, this limit could be useful for you depending on how you approach the problem...)2011-09-03
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    A question on related the $1^{\infty}$ indeterminate form: http://math.stackexchange.com/questions/10490/why-is-1-infty-considered-to-be-an-indeterminate-form2011-09-03
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    I think a more interesting problem would be $\lim_{x \to 0} \Bigl(\frac{\sin{x}}{x}\Bigr)^{1/x^{2}}$2011-09-04
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    @R.M. $\lim_{x\to0^+}(\frac{\sin x}{x})^{x^p}=${$e^{-\frac{1}{6}}$ if $p=-2; 0$ if $p<-2; 1 $ if $ p>-2$}. (It is easy to show by taking $\ln$ and use L'hopital and is obvious from graph https://www.desmos.com/calculator/xrbz1aypnp. For $\lim_{x\to^-}$, by $\frac{\sin (-x)}{-x}=\frac{\sin x}{x}$, it is easy to show your limit doesn't exist2018-05-02

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