<s>[INST] 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?
 [/INST] 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, 
$$
\begin{align}
\lim_{x\to0^-}\left(\left(1+\frac{x^2}3+o\!\left(x^2\right)\right)^{1/x^2}\right)^{1/x}
&=\lim_{x\to0^-}\left(e^{1/3}\right)^{1/x}\\
&=0
\end{align}
$$
</s>