To evaluate $\lim_{x \to 0^-}({\frac{\tan x}{x}})^\frac{1}{x^3}$ Evaluate $$\lim_{x \to 0^-}({\frac{\tan x}{x}})^\frac{1}{x^3}$$
I tried taking log on both sides and then using L'Hospital rule but its giving complex results.Are there any simpler methods to approach this?
 A: From this answer, we can see that the Taylor coefficients of $\tan(x)$ expanded around $0$ will be positive, which implies that every truncation of the Taylor series acts as a lower bound for $\tan(x)$ on $[0, \pi/2)$. Therefore, $$\tan(x)\geq x+\frac{x^3}{3}$$ for $x\in [0, \pi/2)$. As $\tan(x)/x$ is an even function, this implies $$\frac{\tan(x)}{x}\geq 1+\frac{x^2}{3}$$ for $x\in (-\pi/2, 0)\cup (0, \pi/2)$. Therefore, as $1/x^3$ is negative for $x < 0$, $$\left(\frac{\tan(x)}{x}\right)^{1/x^3}\leq \left(1+\frac{x^2}{3}\right)^{1/x^3}$$ By Bernoulli's inequality, $$\left(1+\frac{x^2}{3}\right)^{-1/x^3}\geq 1-\frac{1}{x^3}\cdot \frac{x^2}{3} = 1-\frac{1}{3x}$$ for $-1\leq x < 0$, so $$\left(1+\frac{x^2}{3}\right)^{1/x^3}\leq \left(1-\frac{1}{3x}\right)^{-1} = \frac{3x}{3x-1}$$ Therefore, $$0\leq \lim_{x\to 0^-} \left(\frac{\tan(x)}{x}\right)^{1/x^3}\leq \lim_{x\to 0^-} \left(1+\frac{x^2}{3}\right)^{1/x^3}\leq \lim_{x\to 0^-} \frac{3x}{3x-1} = 0$$
A: By changing $x$ into $-x$ this is the same as
$$
\lim_{x\to0^+}\left(\frac{\tan x}{x}\right)^{-1/x^3}
=
\lim_{x\to0^+}\left(\frac{x}{\tan x}\right)^{1/x^3}
$$
Thus you want to find
\begin{align}
\lim_{x\to0^+}-\frac{\log\dfrac{\tan x}{x}}{x^3}
&=-\lim_{x\to0^+}\frac{\log\left(1+\dfrac{x^2}{3}+o(x^2)\right)}{x^3}\\[6px]
&=-\lim_{x\to0^+}\frac{\dfrac{x^2}{3}+o(x^2)}{x^3}\\[6px]
&=-\infty
\end{align}
Then your limit is $\lim_{t\to-\infty}e^t=0$
A: From Are all limits solvable without L'Hôpital Rule or Series Expansion,
$\lim_{x\to0}\left(\dfrac{\tan x-x}{x^3}\right)=\dfrac13$
$\implies\dfrac{\tan x-x}{x^m}\to0$ for $m<3$ as $x\to0$
$$\lim_{x\to0}\left(\dfrac{\tan x}x\right)^{1/x^3}$$
$$=\left(\left(\lim_{x\to0}\left(1+\dfrac{\tan x-x}x\right)^{x/(\tan x-x)}\right)^{\lim_{x\to0}\frac{\tan x-x}{x^3}}\right)^{\lim_{x\to0}\frac1x}$$
The inner limit converges to $e^{1/3}$ 
What about the outermost exponent? 
A: Equation $(4)$ in this answer says that
$$
\lim_{x\to0}\frac{\tan(x)-x}{x^3}=\frac13
$$
Therefore,
$$
\begin{align}
\left(\frac{\tan(x)}x\right)^{1/x^3}
&=\left(1+\frac{x^2}3+o\!\left(x^2\right)\right)^{1/x^3}\\
&=\left(\left(1+\frac{x^2}3+o\!\left(x^2\right)\right)^{1/x^2}\right)^{1/x}
\end{align}
$$
as $x\to0$, we can make $\left(1+\frac{x^2}3+o\!\left(x^2\right)\right)^{1/x^2}$ as close to $e^{1/3}\gt1$ as we wish. Thus, 
$$
\begin{align}
\lim_{x\to0^-}\left(\left(1+\frac{x^2}3+o\!\left(x^2\right)\right)^{1/x^2}\right)^{1/x}
&=\lim_{x\to0^-}\left(e^{1/3}\right)^{1/x}\\
&=0
\end{align}
$$
A: Easy trick
$$\lim_{x\to 0^-} \left(\frac{\tan x}{x}\right)^{\frac1{x^3}}  =\lim_{x\to 0^-}\exp\left(\frac{1}{x^3}\ln\left(\frac{\tan x -x}{x}+1\right)\right) \sim \lim_{x\to 0^-}\exp\left(\frac{1}{3x}\frac{\ln\left(1+\frac{x^2}{3}\right)}{\frac{x^2}{3}}\right)\\= \color{blue}{\exp(-\infty\times \frac13)= 0} $$
Given that $$\tan x -x \sim \frac{x^3}{3}~~~~and ~~~~ \lim_{h\to 0} \frac{\ln\left(1+h\right)}{h} = 1$$
