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?

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, $$0\leq \lim_{x\to 0^-} \left(\frac{\tan(x)}{x}\right)^{1/x^3}\leq \lim_{x\to 0^-} \left(1+\frac{x^2}{3}\right)^{1/x^3}\leq \lim_{x\to 0^-} \frac{3x}{3x-1} = 0$$
