How to Find $ \lim\limits_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$. Can someone help me with this limit? I'm working on it for hours and cant figure it out.
$$ \lim_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$$
I started transforming to the form
$ \lim_{x\to 0} e^{  {\frac{\ln \left(\frac {\tan x}{x} \right)}{x^2}}  }$
and applied the l'Hopital rule (since indeterminated $\frac00$), getting:
$$ \lim_{x\to 0} \left( \frac{2x-\sin 2x }{2x^2\sin 2x} \right)$$
From here, I try continue with various forms of trigonometric substitutions, appling the l'Hopital rule again and again, but no luck for me. Can someone help me?
 A: Whenever we have an expression where both base and exponent are variables, it is best to take logs. Thus if $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{x \to 0}\left(\frac{\tan x}{x}\right)^{1/x^{2}}\right\}\notag\\
&= \lim_{x \to 0}\log\left(\frac{\tan x}{x}\right)^{1/x^{2}}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(\frac{\tan x}{x}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\frac{\tan x - x}{x}\cdot\dfrac{\log\left(1 + \dfrac{\tan x - x}{x}\right)}{\dfrac{\tan x - x}{x}}\notag\\
&= \lim_{x \to 0}\frac{\tan x - x}{x^{3}}\cdot 1\notag\\
&= \lim_{x \to 0}\frac{\sec^{2} x - 1}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{\tan^{2}x}{x^{2}}\notag\\
&= \frac{1}{3}\notag
\end{align}
Hence $L = e^{1/3}$.
A: Put 
$$ y = \left({\tan(x)\over x}\right)^{1/x^2}.$$
Then $$\log(y) = {\log(\tan(x)) - \log(x)\over x^2}$$
Can you evaluate the RHS with the help of L'hospital?
A: L'Hospital's rule is not the alpha and omega of limits computation! 
As $\tan x= x+\dfrac{x^3}3+o(x^3)$,
$$\frac{\tan x}x=1+\frac{x^2}3+o(x^2),\enspace\text{hence}\enspace \frac 1{x^2}\ln\Bigl(\frac{\tan x}x\Bigr)=\frac 1{x^2}\ln\Bigl(1+\frac{x^2}3+o(x^2)\Bigr)=\frac 13+o(1)\to
 \frac 13$$
so that $$\lim_{x\to 0}\Bigl(\frac{\tan x}x\Bigr)^{\tfrac 1{x^2}}=\mathrm e^{\frac 13}.$$
A: Let $\displaystyle y(x)
=\ln\left[\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}\right]
=\frac{1}{x^2}\ln\left(\frac{\tan x}{x}\right)$, then by using the 
L'Hôpital's rule several times,
\begin{align}
\lim_{x\to 0}y(x)&=\lim_{x\to 0}\frac{\ln\left(\frac{\tan x}{x}\right)}{x^2}\\
&\stackrel{{\rm H}}{=}
\lim_{x\to 0}\frac{\frac{x}{\tan x}\cdot\frac{x\sec^2x-\tan x}{x^2}}{2x}\tag{1}\\
&=\lim_{x\to 0}\frac{x\sec^2x-\tan x}{2x^2\tan x}\\
&\stackrel{{\rm H}}{=}
\lim_{x\to 0}\frac{\sec^2x+2x\sec^2x\tan x-\sec^2 x}{4x\tan x+2x^2\sec^2x}\\
&=\lim_{x\to 0}\frac{2\sec^2x\tan x}{4\tan x+2x\sec^2x}\\
&=\lim_{x\to 0}\frac{2\sec^2x}{4+2\left(\frac{x}{\sin x}\right)\sec x}\\
&=\frac{2\lim_{x\to 0}\sec^2x}{4+2\left(\lim_{x\to 0}\frac{x}{\sin x}\right)\left(\lim_{x\to 0}\sec x\right)}\\
&=\frac{1}{3}.
\end{align}
Hence $\displaystyle
\lim_{x\to0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}
=\lim_{x\to0}e^{y(x)}
=e^{\lim_{x\to0}y(x)}=e^{\frac{1}{3}}$. Notice that $(1)$ holds because
$$\lim_{x\to0}\ln\left(\frac{\tan x}{x}\right)
=\ln\left(\lim_{x\to 0}\frac{\tan x}{x}\right)
\stackrel{{\rm H}}{=}
\ln\left(\lim_{x\to 0}\frac{\sec^2 x}{1}\right)=\ln(1)=0,$$
which follows that $\displaystyle \frac{\ln\left(\frac{\tan x}{x}\right)}{x^2}\rightarrow\frac{0}{0}$ and we can use the L'Hôpital's rule.
A: It is possible to compute this limit only by making use of some basic limits:
\begin{eqnarray*}
\lim_{u\rightarrow 0}\frac{\ln (1+u)}{u} &=&1 \\
\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}} &=&\frac{1}{3} \\
\lim_{x\rightarrow 0}\frac{\tan x}{x} &=&1.
\end{eqnarray*}
Take the logarithm
\begin{eqnarray*}
L &=&\ln \left( \frac{\tan x}{x}\right) ^{1/x^{2}} \\
&=&\frac{1}{x^{2}}\ln \frac{\tan x}{x} \\
&=&\frac{1}{x^{2}}\ln \left( 1+\frac{\tan x}{x}-1\right)  \\
&=&\frac{\left[ \frac{\tan x}{x}-1\right] }{x^{2}}\cdot \frac{\ln \left( 1+%
\left[ \frac{\tan x}{x}-1\right] \right) }{\left[ \frac{\tan x}{x}-1\right] }
\\
&=&\frac{\tan x-x}{x^{3}}\cdot \frac{\ln (1+u(x))}{u(x)},\ with\ u(x)=\frac{%
\tan x}{x}-1
\end{eqnarray*}
since
\begin{equation*}
\lim_{x\rightarrow 0}u(x)=\lim_{x\rightarrow 0}(\frac{\tan x}{x}-1)=1-1=0
\end{equation*}
then
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=\lim_{u\rightarrow 0}\frac{%
\ln (1+u)}{u}=1,
\end{equation*}
and therefore
\begin{equation*}
\lim_{x\rightarrow 0}L(x)=\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}\cdot
\lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=\frac{1}{3}\cdot 1=1.
\end{equation*}
Backward, one take the exponential, and using continuity arguments, obtains
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{\tan x}{x}\right)
^{1/x^{2}}=\lim_{x\rightarrow 0}e^{L(x)}=e^{1/3}.
\end{equation*}
A: If we are allowed to use Series Expansion,
$\tan x=x+\dfrac{x^3}3+O(x^5)$
$$\lim_{x\to0}\left(\dfrac{\tan x}x\right)^{1/x^2}=\left(\lim_{x\to0}\left(1+\dfrac{x^2}3+O(x^4)\right)^{1/(1+x^2/3+O(x^4))}\right)^{\lim_{x\to0}\dfrac{1+x^2/3+O(x^4)}{x^2}}$$
Set $\dfrac1{1+x^2/3+O(x^4)}=n$ in the inner limit
Can you take it from here!
A: If $\lim_{x\rightarrow a}f(x)=1$ and $\lim_{x\rightarrow a}g(x)=\infty$, then,
$$\lim_{x\rightarrow a}f(x)^{g(x)}=e^{\lim_{x\rightarrow a}(f(x)-1)g(x)}$$
This is because:
Let $f$ and $g$ be functions such that $\lim_{x\rightarrow a}f=1$ and $\lim_{x\rightarrow a}g=\infty$.
Let $$L=\lim_{x\rightarrow a}f^g$$
$$\log L=\log(\lim_{x\rightarrow a}f^g)$$
$$\log L=\lim_{x\rightarrow a}\log(f^g)$$
$$\log L=\lim_{x\rightarrow a}g\log(f)$$
$$L=e^{\lim_{x\rightarrow a}g\log(f)}$$
$$L=e^{\lim_{x\rightarrow a}g\frac{\log(f)}{f-1}(f-1)}$$
$$L=e^{\lim_{x\rightarrow a}g\lim_{x\rightarrow a}\frac{\log(f)}{f-1}\lim_{x\rightarrow a}(f-1)}$$
$$\lim_{x\rightarrow a}\frac{\log(f)}{f-1}=1$$
Thus,
$$L=e^{\lim_{x\rightarrow a}g(f-1)}$$
Here,
$$ \lim_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}=\lim_{x\to 0} e^{\left(\frac {\tan x }{x} -1\right){\frac{1}{x^2}}}=\lim_{x\to 0} e^{\left(\frac {\tan x  -x}{x^3}\right)}=\lim_{x\to 0} e^{\left(\frac {x+\frac{x^3}{3}+O(x^5)  -x}{x^3}\right)}=e^{\frac13}$$
A: In the same spirit as other answers, consider $$A=\left(\frac {\tan (x) }{x} \right)^{\frac{1}{x^2}}$$ Take logarithms $$\log(A)=\frac{1}{x^2}\log\left(\frac {\tan (x) }{x} \right)$$ Now, consider Taylor series $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ $$\frac {\tan (x) }{x}=1+\frac{x^2}{3}+\frac{2 x^4}{15}+O\left(x^6\right)$$ $$\log\left(\frac {\tan (x) }{x} \right)=\log\left(1+\frac{x^2}{3}+\frac{2 x^4}{15}+O\left(x^6\right)\right)$$  Now use $\log(1+y)=y-\frac{1}{2}y^2+O\left(y^3\right)$ and replace $y$ by $\frac{x^2}{3}+\frac{2 x^4}{15}$ to get $$\log(A)=\frac{1}{x^2} \left(\frac{x^2}{3}+\frac{7 x^4}{90}+O\left(x^5\right)\right)=\frac{1}{3}+\frac{7 x^2}{90}+O\left(x^3\right)$$ which shows the limit and also how it is approached. 
You can continue using $A=e^{\log(A)}$ and still using Taylor arrive to $$A=\sqrt[3]{e}+\frac{7}{90} \sqrt[3]{e} x^2+O\left(x^3\right)$$
A: It is in the standard form of $1^{\infty}$ 
The general value of these type of limits is $e^{\lim_{ x \rightarrow 0}{(f(x)-1)g(x)}}$
Here $f(x)=\frac{\tan x}{x},g(x)=\frac{1}{x^2}$
Therefore you need to find $\lim_{x \rightarrow 0} \frac{\tan x-x}{x^3}$ 
Now, apply L Hospital to get $\lim_{x \rightarrow 0} \frac{ \tan x -x}{x^3}=\frac{ \tan ^2 x}{3x^2}=\frac{1}{3}$
Thus the limit is $e^{\frac{1}{3}}$
A: $$\lim_{x\to 0} \left(\frac{\tan x}{x}\right)^{\frac1{x^2}}  =\lim_{x\to 0}\exp\left(\frac{1}{x^2}\ln\left(\frac{\tan x -x}{x}+1\right)\right) \sim \lim_{x\to 0}\exp\left(\frac{1}{3}\frac{\ln\left(1+\frac{x^2}{3}\right)}{\frac{x^2}{3}}\right)= \color{blue}{\exp(\frac13)}$$
Given that $$\tan x -x \sim \frac{x^3}{3}~~~~and ~~~~ \lim_{h\to 0} \frac{\ln\left(1+h\right)}{h} = 1$$
