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I have to find: $\lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}}$

"Minimized" the problem to finding: $ \lim_{x \to 0^+} \frac {\ln(\frac{\arctan x}{x})} {x^2} $

L'Hôpital's rule once is not enough, second L'Hôpital's rule seems worse. any help?

Edit: I cannot use power series expansion. that is provided on the next semester.

user7610
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4 Answers4

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Let $\displaystyle A=\lim_{x\to0}\frac{\arctan x-x}{x^3}$

Applying L'Hospital's Rule $\displaystyle A=\lim_{x\to0}\frac{\dfrac1{1+x^2}-1}{3x^2}=-\lim_{x\to0}\frac{x^2}{3x^2(1+x^2)}=-\frac13$

So, $\displaystyle \lim_{x\to0^+}\left(\frac{\arctan x}x\right)^{\dfrac1{x^2}}=\lim_{x\to0^+}\left(1+\frac{\arctan x-x}{x^3}\cdot x^2\right)^{\dfrac1{x^2}}$

$\displaystyle \approx \lim_{x\to0^+}\left(\left(1+\frac{-x^2}3\right)^{\dfrac3{-x^2}}\right)^{-\dfrac13}$

Now we know $\displaystyle\lim_{h\to0}\left(1+h\right)^{\dfrac1h}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$

lab bhattacharjee
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To add to lab bhattacharjee-s answer:

From $ \lim_{x \to 0} \frac{\log (1+x)}{x} = 1 $ (which you can easily derive using L'Hôpital's rule for example): $$ \lim_{x \to 0^+} \frac{\log \frac{\arctan x}{x}}{x^2} = \lim_{x \to 0^+} \frac{\log(1+ \frac{\arctan x-x}{x})}{\frac{\arctan x-x}{x}} {\frac{\arctan x-x}{x^3}}$$ (you can check this by actually doing the cancellations) $$ = \lim_{x \to 0^+} \frac{\log(1+ \frac{\arctan x-x}{x})}{\frac{\arctan x-x}{x}} \lim_{x\to0^+}{\frac{\arctan x-x}{x^3}} = -\frac13 $$ (by the above identity and L'Hôpital's rule, as you can see in lab bhattacharjee-s answer)

It follows that the answer for your question is $e^{-\frac{1}{3}}$ (since $ \frac{\log \frac{\arctan x}{x}}{x^2} $ is the logarithm of the original limit). I hope that's complete enough!

Dániel G.
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  • Beautiful answer! – user7610 Jan 22 '14 at 10:46
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    to readers: I'm accepting this answer simply for it's unexpectedness feel. for a thorough, simpler and technical answer see below Americo's answer. – user7610 Jan 23 '14 at 00:02
  • Isn't $\lim_{x\to 0} \frac{\log(1+x)}{x}=1$ is the definition of the derivative of $log(1+x)$ at $x=0$? It's a bit weird to prove it using L'hopital that uses the derivative of $\log(1+x)$, no? – Lior B-S Jan 28 '14 at 05:35
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"Minimized" the problem to finding: $ \lim_{x \to 0^+} \frac {\ln(\frac{\arctan x}{x})} {x^2}$

L'Hôpital's rule once is not enough, second L'Hôpital's rule seems worse. any help?

By applying L'Hôpital's rule trice and simplifying in each step we obtain:

\begin{eqnarray*} \lim_{x\rightarrow 0^{+}}\frac{\ln \left( \frac{\arctan x}{x}\right) }{x^{2}} &=&\lim_{x\rightarrow 0^{+}}\frac{\ln \left( \arctan x\right) -\ln \left( x\right) }{x^{2}}\tag{1} \\ &=&\frac{1}{2}\lim_{x\rightarrow 0^{+}}\frac{x-\arctan x-x^{2}\arctan x}{ x^{2}\arctan x+x^{4}\arctan x}\tag{2} \\ &=&\lim_{x\rightarrow 0^{+}}\frac{-\arctan x}{4x^{2}\arctan x+x+2\arctan x} \tag{3}\\ &=&\lim_{x\rightarrow 0^{+}}\frac{-1}{5x^{2}+8x\arctan x+8x^{3}\arctan x+3} \tag{4}\\ &=&-\frac{1}{3}.\tag{5} \end{eqnarray*}

ADDED. Explanation:
$\ \:(1)$: $\frac{d}{dx}\left( \ln \left( \arctan x\right) -\ln \left( x\right) \right) =\dfrac{x-\arctan x-x^{2}\arctan x}{x\left( 1+x^{2}\right) \arctan x}$;

$\ \:(2)$: $\frac{d}{dx}\left( x-\arctan x-x^{2}\arctan x\right) =-2x\arctan x$;

$\ \:(2)$: $\frac{d}{dx}\left( x^{2}\arctan x+x^{4}\arctan x\right) =4x^{3}\arctan x+x^{2}+2x\arctan x$;

$\ \:(3)$: $\frac{d}{dx}\left( -\arctan x\right) =-\dfrac{1}{1+x^{2}}$;

$\ \:(3)$: $\frac{d}{dx}\left( 4x^{2}\arctan x+x+2\arctan x\right) =\dfrac{5x^{2}+8x\arctan x+8x^{3}\arctan x+3}{1+x^{2}}$.

Américo Tavares
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\begin{align*} \lim_{x \to 0^+} (\frac{\arctan (x)}{x})^{\frac{1}{x^2}}&=\lim_{x \to 0^+} (\frac{x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} +... }{x})^{\frac{1}{x^2}}\\ &=\lim_{x \to 0^+}(1-\frac {x^2}3+ \frac{x^4}{5} - \frac{x^6}{7} +...)^{1/x^2}\\ &=e^{-1/3} \end{align*}

Jlamprong
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