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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?

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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?

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  • $\begingroup$ Isn't it $e^0$? $\endgroup$ – PiGamma Nov 20 '17 at 6:19
  • $\begingroup$ $\frac{x}{\tan(x)-x}\cdot \frac{\tan(x)-x}{x^2} = \frac{1}{x}$, not $\frac{1}{x^3}$. $\endgroup$ – Michael Lee Nov 20 '17 at 6:19
  • $\begingroup$ Actually, $\lim_{x\to 0} \left(\frac{\tan(x)}{x}\right)^{1/x^3}$ does not exist. From the left-hand side, it's $0$, and from the right-hand side, it's infinite. You can see as much by plotting the function using WolframAlpha. $\endgroup$ – Michael Lee Nov 20 '17 at 6:20
  • $\begingroup$ @MichaelLee I wanted to calculate only left hand limit $\endgroup$ – PiGamma Nov 20 '17 at 6:22
  • $\begingroup$ There's a mistake in your notation, then. You wrote $x\to 0-0$ where you meant $x\to 0^-$. I've just edited it. $\endgroup$ – Michael Lee Nov 20 '17 at 6:23
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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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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 $\lim_{t\to-\infty}e^t=0$

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

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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 $$\tan x -x \sim \frac{x^3}{3}~~~~and ~~~~ \lim_{h\to 0} \frac{\ln\left(1+h\right)}{h} = 1$$

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