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About the limit $\displaystyle\lim_{x\to \infty} \left(\frac{\sin x}x\right)^{1/{x^2}}$

Finding it: $$\lim_{x\to \infty}\exp\left(\frac{\ln \frac{\sin x}{x}}{x^2}\right)=\lim_{x\to \infty}\exp\left(\frac{\ln 1}{x^2}\right)=\exp(0)=1$$

But when I place various numbers in $\left(\frac{\sin x}x\right)^{1/{x^2}}$ I get many times results in complex numbers so I'm starting to doubt if my calculation is correct. Can anyone shed some light on this please?

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  • $\begingroup$ Please refrain from using \left and \right in titles because it causes issues with the HTML preview being much too tall. $\endgroup$ – AlexR Feb 10 '15 at 19:04
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    $\begingroup$ We have $$\lim_{x\nearrow (2k+1)\pi} \biggl(\frac{\sin x}{x}\biggr)^{1/x^2} = 0$$ for every $k\in\mathbb{N}$. On the other hand, for $a_k = (2k+1/2)\pi$, we have $$\biggl(\frac{\sin a_k}{a_k}\biggr)^{1/a_k^2} = \frac{1}{a_k^{1/a_k^2}} \to 1.$$ So the limit you're looking for does not exist. $\endgroup$ – Daniel Fischer Feb 10 '15 at 20:17
  • $\begingroup$ So rubik's answer is actually wrong? @DanielFischer $\endgroup$ – GinKin Feb 10 '15 at 20:36
  • $\begingroup$ Indeed, the answers here aren't correct. $\endgroup$ – Daniel Fischer Feb 10 '15 at 20:42
  • $\begingroup$ @DanielFischer if the function would be defined such that $\sin x >0, \forall x$ then will the expression have a limit? $\endgroup$ – GinKin Feb 10 '15 at 21:09
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The limit of $f(x) = \bigl(\frac{\sin x}{x}\bigr)^{1/x^2}$ as $x \to \infty$ does not exist. On the one hand, at the zeros of $\sin$ we have zeros of $f$ - at $x_k = k\pi,\; k\in \mathbb{Z}$, the expression becomes $0^{1/(k\pi)^2}$, and $\frac{1}{(k\pi)^2} > 0$ - so

$$\lim_{k\to\infty} f(x_k) = 0,$$

and on the other hand for $a_k = (2k+\frac{1}{2})\pi,\; k\in\mathbb{Z}$ we have

$$f(a_k) = \biggl(\frac{\sin a_k}{a_k}\biggr)^{1/a_k^2} = a_k^{-1/a_k^2},$$

which gives us

$$\lim_{k\to\infty} f(a_k) = 1.$$

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$\frac{\sin x}x\to 0$ as $x\to\infty$, not 1.

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Correct. Because of the sine function the results are highly oscillating.

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