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