L'Hospital Rule With Exponents - $ \tan x/x $ 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 ! 
 A: 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 
$$\lim_{x\to 0} \left(\frac{\tan x}x\right) ^{x^{-2}} = \sqrt[3]{e}.$$
A: $$ \lim _ {x \to 0 } \left (\frac{\tan x } {x}  \right ) ^{1/x^2}=\lim_{x\to 0}\left(1+\left({\frac{\tan x-x}{x}}\right)\right)^{{x\over\tan x-x}{\tan x-x\over x^3}}=e^{\lim_{x\to 0}\frac{\tan x-x}{x^3}}$$
$$\lim_{x\to 0}\frac{\tan x-x}{x^3}=\lim_{x\to 0}\frac{{1\over \cos^2x}-1}{3x^2}={1\over 3}\lim_{x\to 0}\left({\tan x\over x}\right)^2={1\over 3}$$
$$ \lim _ {x \to 0 } \left (\frac{\tan x } {x}  \right ) ^{1/x^2}=e^{1/3}$$
A: 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 $$\lim_{x\to+\infty}\exp\left(\frac{\tan(x)-x}{x^3}\right)=\exp(1/3)$$
A: 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. 
A: 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)
$$=\frac8{8+2\cdot8}=\frac13$$
