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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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Rewriting as $e^{\ln ...}$ is a good idea. Maybe you should apply l'Hospital it twice. – Fabian Dec 31 '12 at 16:34
applying it twice also doesn't help ... So my guess is that there is some other way to do it . – theMissingIngredient Dec 31 '12 at 16:35
There is some other way to do it. Do you know Taylor expansion? – Fabian Dec 31 '12 at 16:36
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 ! – theMissingIngredient Dec 31 '12 at 16:45

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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thanks a lot !!!!!!!!!! – theMissingIngredient Dec 31 '12 at 17:10
just one thing: in your 9th line, you wrote $ ln (1+\frac{x^2 }{3} +O(x^4) $ = $\frac{x^2 }{3} +O(x^4) $ . Shouldn't you also change something in the big-o ? thanks!@ – theMissingIngredient Dec 31 '12 at 17:20
i.e.- your $y$ is $ y = \frac{x^2}{3} +O(x^4 ) $ , and you are looking at $ ln(1+y)$ – theMissingIngredient Dec 31 '12 at 17:20
The next term in the expansion of $\ln (1+y)= x^2/3+ ...$ comes from the first order term of $O(x^4)$ and the second order term of $x^2/3$ both of which are of order $O(x^4)$. (hope this is understandable) – Fabian Dec 31 '12 at 17:25
Soryy, but I couldn't understand ... Shouldn't you have $O(x^8)$ after substituting? – theMissingIngredient Dec 31 '12 at 18:37

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

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

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How good do you see my approach, Adi?:) – Babak S. Jan 1 '13 at 7:35

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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Helpful to know! +1 – amWhy Feb 26 '13 at 0:54
@amWhy: Thanks ;-) – Babak S. Feb 26 '13 at 2:58
I really like how you contribute, giving diverse ways of looking at problems, and doing so in a helpful, nudging, and instructive way. Students so often fall into the "trap" of thinking there is only one correct answer, or one way of doing things, you help bring "creativity" to math! – amWhy Feb 26 '13 at 3:01
Thanks for saying so. It is very kind of you @amWhy. :-) – Babak S. Feb 26 '13 at 3:03

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)


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wow ! great ! thanks a lot !!! – theMissingIngredient Dec 31 '12 at 18:40
@theMissingIngredient, my pleasure. I think the problem is best solved by Fabian's way if 'L'Hospital's Rule' is not mentioned. – lab bhattacharjee Dec 31 '12 at 18:43

One might rewrite @Fabian's answer without big-O thus:

Since $\tan x = x + {x^3\over3} + {2x^5\over15}+\cdots$ and $\log(1+x) = x - {x^2\over2} + \cdots$, we have $$\left({\tan x\over x}\right)^{1/x^2} =e^{{1\over x^2}\log\left({\tan x \over x}\right)} =e^{{1\over x^2}\log\left(1+{x^2\over3}+{2x^4\over15}+\cdots\right)} =e^{{1\over x^2}\left(\left({x^2\over3}+{2x^4\over15}+\cdots\right)-{1\over2}\left({x^2\over3}+{2x^4\over15}+\cdots\right)^2+\cdots\right)} =e^{{1\over3}+{7x^2\over90}+\cdots}$$ which approaches $\root 3 \of e$.

However in a calculus course based on standard US textbooks, the problem looks like a standard, if involved, L'Hopital's Rule problem, which can be handled as @lab bhattacharjee, with one simplification thus (all limits as $x\rightarrow0$): $$\log\lim \left({\tan x\over x}\right)^{1/x^2} = \lim {\log\left({\tan x\over x}\right)\over{x^2}} \buildrel L'H \over = \lim {2x -\sin 2x \over 2x^2 \sin2x} = \lim {2x -\sin 2x \over 4x^3} \lim {2x \over \sin2x} \buildrel L'H \over = \lim {2\over3}{1 -\cos 2x \over (2x)^2} = {1 \over3}\,, $$ the last limit either being known to be $1/2$ or done by L'Hopital's rule twice.

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Great !! Thanks a lot !!! – theMissingIngredient Jan 1 '13 at 21:02

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