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given the following limit:

$ \lim _ {x \to 0 } \left (\frac{\tan x } {x} \right ) ^{1/x^2}\;, $

is there any simple way to calculate it ?

I have tried writing it as $e^ {\ln (\dots)} $ , but it doesn't give me anything [ and I did l'Hospital on the limit I received... ]

can someone help me with this?

thanks a lot !

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    Yes, but I am pretty sure we need to do it with l'Hospital Is there any trick to solve it with l'Hospital ? Thanks !2012-12-31

5 Answers 5

5

If you know Taylor expansion, you know that $\tan x = x + \frac{x^3}{3}+ \mathcal{O}(x^5)$ where the big-Oh denotes a term which scales like $x^5$ for $x\to 0$. Thus, $\frac{\tan x}{x} = 1 + \frac{x^2}{3} + \mathcal{O}(x^4).$ The expansion of the logarithm around $1$ reads $ \ln (1+y) = y + \mathcal{O}(y^2).$ Letting $1+y=\tan x/x =1+ x^2/3 + \mathcal{O}(x^4)$, we obtain $ \ln \left(\frac{\tan x}x \right) = \frac{x^2}3 + \mathcal{O}(x^4).$ Now, $ \frac1 {x^2} \ln \left(\frac{\tan x}x \right) = \frac13 + \mathcal{O}(x^2).$ And thus $\lim_{x\to 0} \frac1 {x^2} \ln \left(\frac{\tan x}x \right) = \frac13.$

With that you can easily show that $\lim_{x\to 0} \left(\frac{\tan x}x\right) ^{x^{-2}} = \sqrt[3]{e}.$

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    got it ! thanks !!!2013-01-01
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$ \lim _ {x \to 0 } \left (\frac{\tan x } {x} \right ) ^{1/x^2}=\lim_{x\to 0}\left(1+\left({\frac{\tan x-x}{x}}\right)\right)^{{x\over\tan x-x}{\tan x-x\over x^3}}=e^{\lim_{x\to 0}\frac{\tan x-x}{x^3}}$

$\lim_{x\to 0}\frac{\tan x-x}{x^3}=\lim_{x\to 0}\frac{{1\over \cos^2x}-1}{3x^2}={1\over 3}\lim_{x\to 0}\left({\tan x\over x}\right)^2={1\over 3}$ $ \lim _ {x \to 0 } \left (\frac{\tan x } {x} \right ) ^{1/x^2}=e^{1/3}$

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    How good do you see my approach, Adi?:)2013-01-01
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Use this fact as well:

If $\lim\limits_{x\to{+\infty}} f(x)^{g(x)}$ be as $1^{+\infty}$, which is an indeterminate form, then we have this fact that: $\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim\limits_{x\to +\infty}\big(f(x)-1\big)g(x)}$

Here we have $\lim_{x\to+\infty}\exp\left(\frac{\tan(x)-x}{x^3}\right)=\exp(1/3)$

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    Thanks for saying so. It is very kind of you @amWhy. :-)2013-02-26
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It is possible to do it with l'Hospital's rule. It takes 4 applications, but it does work! Do the exponential transformation, and continue simplifying with l'Hospital's rule and limits until you get:

$ \exp\left(-\frac{\left(2 \left(\lim_{x\rightarrow0} \cos\left(2 x\right)\right)\right)}{\left(\lim_{x\rightarrow0} \left(\left(4 x^2-6\right) \cos\left(2 x\right)+12 x \sin\left(2 x\right)\right)\right)}\right) $

Use a bunch of limit rules (quotient, continuity, sum, product, polynomial, in that order), in order to get: $ \exp\left(-\frac{\left(2 \cos\left(\lim_{x\rightarrow0} 2 x\right)\right)}{\left(12 \left(\lim_{x\rightarrow0} x\right) \left(\lim_{x\rightarrow0} \sin\left(2 x\right)\right)-6 \cos\left(\lim_{x\rightarrow0} 2 x\right)\right)}\right) $

Just evaluate all the limits now: $ \exp\left(\frac{1}{3}\right) $

Tedious, but certainly doable! I would recommend Fabian's solution instead, 4 applications of Hospital's, although possible, is something you want to avoid if possible.

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Let $y=\left (\frac{\tan x } {x} \right ) ^{1/x^2}$

So, $\ln y=\frac{\ln \tan x -\ln x}{x^2}$

Then $\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\ln \tan x -\ln x}{x^2}\left(\frac 00\right)$ as $\lim_{x\to 0}\frac {\tan x}x=1$

Applying L'Hospital's Rule: , $\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\frac2{\sin2x}-\frac1x}{2x}=\lim_{x\to 0}\frac{2x-\sin2x}{2x^2\sin2x}\left(\frac 00\right)$

$=\lim_{x\to 0}\frac{2-2\cos2x}{4x\sin2x+4x^2\cos 2x}\left(\frac 00\right)$ (applying L'Hospital's Rule)

$=\lim_{x\to 0}\frac{4\sin2x}{4\sin2x+8x\cos2x+8x\cos 2x-8x^2\sin2x}\left(\frac 00\right)$ (applying L'Hospital's Rule)

$=\lim_{x\to 0}\frac{8\cos2x}{8\cos2x+2(8\cos2x-16x\sin2x)-16x\sin2x-16x^2\cos2x}\left(\frac 00\right)$ (applying L'Hospital's Rule)

$=\frac8{8+2\cdot8}=\frac13$

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    @theMissingIngredient, my pleasure. I think the problem is best solved by Fabian's way i$f$ 'L'Hospital's Rule' is not mentioned.2012-12-31