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February 17th, 2017, 01:50 AM   #1
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Limit problem!

What is the value of
(Sinx/x)^1/(x^2)
Where x tends to 0?
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February 17th, 2017, 03:40 AM   #2
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Is that $\displaystyle \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{x^2}}$?

As a first thought, have you tried a series expansion of $\sin x$ or l'Hôpital's rule?

Last edited by v8archie; February 17th, 2017 at 03:50 AM.
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February 17th, 2017, 02:43 PM   #3
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Try taking log. I get limit = $\displaystyle e^{-\frac{1}{6}}$.
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February 17th, 2017, 02:57 PM   #4
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That's what the power series $\displaystyle \sin x = x-\tfrac{1}{3!}x^3 + \mathcal{O}(x^5)$ would suggest - and rather more easily than l'Hôpital's rule. (I did it in my head.)

Last edited by v8archie; February 17th, 2017 at 02:59 PM.
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February 17th, 2017, 06:25 PM   #5
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Can you elaborate on how you did the power series in your head to come up with $\displaystyle e^{-\frac{1}{6}}$, please? How did you do the power series of it when it's being raised to by $\displaystyle \frac{1}{x^2}$?

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February 17th, 2017, 07:22 PM   #6
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There is a standard result that you should learn and remember (it is a definition of $e$):
$$\lim_{x \to \infty} \left(1 + \frac{1}{x} \right)^{x} = e$$
If we then set $y = \frac{1}{x}$ we get $y \to 0$ as $x \to \infty$ (there's a small shortcut in that statement, but don't worry about it) and then
$$\lim_{y \to 0} \left(1 + y\right)^{\frac{1}{y}} = e$$

The power series is something you might remember (or look up). It gives you
\begin{align*}
\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{x^2}} &= \lim_{x \to 0} \left( \frac{x - \frac{1}{3!}x^3 + \mathcal{O}(x^5)}{x} \right)^{\frac{1}{x^2}} \\
&= \lim_{x \to 0} \left( 1 - \tfrac{1}{3!}x^2 + \mathcal{O}(x^4) \right)^{\frac{1}{-\frac{1}{3!}x^2 + \mathcal{O}(x^4)} \cdot \frac{-\frac{1}{3!}x^2 + \mathcal{O}(x^4)}{x^2}} \\
&= \lim_{x \to 0} \left( \left( 1 - \tfrac{1}{3!}x^2 + \mathcal{O}(x^4) \right)^{\frac{1}{-\frac{1}{3!}x^2 + \mathcal{O}(x^4)}} \right)^{-\frac{1}{3!} + \mathcal{O}(x^2)} \\
&= \left( e \right)^{-\frac{1}{6} + 0} \\
&= e^{-\frac16}
\end{align*}
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February 17th, 2017, 09:17 PM   #7
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@V8archie
Thanks!
I got it!
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February 18th, 2017, 02:36 PM   #8
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Using log.
$\displaystyle f(x)=(\frac{\sin x}{x})^{\frac{1}{x^2}}$
$\displaystyle \ln(f(x))=\frac{1}{x^2}\ln(\frac{\sin x}{x})$
$\displaystyle \ln(\frac{\sin x}{x})\approx \ln(1-\frac{x^2}{6})\approx -\frac{x^2}{6}$
Therefore:
$\displaystyle \ln(f(x))\approx -\frac{1}{6}$
or $\displaystyle f(x)\approx e^{-\frac{1}{6}}$
Thanks from topsquark

Last edited by skipjack; February 18th, 2017 at 09:00 PM.
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