Indeterminate Form!!

$\begin{aligned} L & = \lim_{x \to 0} \left(\frac {\tan x}x\right)^{\frac 1{x^2}} & \small \color{#3D99F6} \text{Since }\lim_{x \to 0} \frac {\tan x}x \text{, this is a }1^\infty \text{ case.} \\ & = \exp \left(\lim_{x \to 0} \frac 1{x^2} \left(\frac {\tan x}x - 1\right) \right) & \small \color{#3D99F6} \text{See note.} \\ & = \exp \left(\lim_{x \to 0} \frac {{\color{#3D99F6}\tan x}-x}{x^3} \right) & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \exp \left(\lim_{x \to 0} \frac {{\color{#3D99F6}x + \frac {x^3}3+\frac {2x^5}{15}+\frac {17x^7}{315}+\cdots}-x}{x^3} \right) \\ & = \exp \left(\lim_{x \to 0} \left(\frac 13+\frac {2x^2}{15}+\frac {17x^4}{315}+\cdots \right) \right) \\ & = e^\frac 13 = \boxed{\sqrt[3]e} \end{aligned}$

Note: If $\displaystyle \lim_{x \to a} \left(\frac {f(x)}{g(x)} \right)^{h(x)} = 1^\infty \implies \lim_{x \to a} \left(\frac {f(x)}{g(x)} \right)^{h(x)} = e^{\lim_{x \to a} \frac 1{h(x)}\left(\frac {f(x)}{g(x)} - 1\right)}$ ( See 2nd Method ).

$\lim_{x\to0}\frac{\tan x}{x}=1\qquad\qquad\lim_{x\to0}\frac{1}{x^2}=\infty\\ \lim_{x\to0}\left(\frac{\tan x}{x}\right)^\frac{1}{x^2}=1^\infty$ This is a good example of a case where the indeterminate form $1^\infty$ does not equal $1$ .

First we will take the natural logarithm of the limit to get rid of the power: $\begin{aligned}\ln\left[\lim_{x\to0}\left(\frac{\tan x}{x}\right)^\frac{1}{x^2}\right]&=\lim_{x\to0}\left[\ln\left(\frac{\tan x}{x}\right)^\frac{1}{x^2}\right]\\ &=\lim_{x\to0}\left[\frac{1}{x^2}\ln\left(\frac{\tan x}{x}\right)\right]\\ &=\lim_{x\to0}\frac{\ln\left(\frac{\tan x}{x}\right)}{x^2}\end{aligned}$ Since $\ln(1)=0$ we now have a $\frac{0}{0}$ situation, so we can use L'Hospital's Rule and derive the top and bottom of the fraction to remove the logarithm: $\begin{aligned}&=\lim_{x\to0}\frac{\frac{d}{dx}\ln\left(\frac{\tan x}{x}\right)}{\frac{d}{dx}x^2}\\ &=\lim_{x\to0}\frac{\left(\frac{\left(\frac{x\sec^2x-\tan x}{x^2}\right)}{\left(\frac{\tan x}{x}\right)}\right)}{2x}\\ &=\lim_{x\to0}\frac{x\sec^2x-\tan x}{2x^2\tan x}\end{aligned}$ The top and bottom still equal zero, so we just have to use L'Hospital's Rule again. Remember that $\frac{d}{dx}\tan x=\sec^2x$ and $\frac{d}{dx}\sec^2 x=2\sec^2x\tan x$ . $\begin{aligned}&=\lim_{x\to0}\frac{\frac{d}{dx}\left(x\sec^2x-\tan x\right)}{\frac{d}{dx}\left(2x^2\tan x\right)}\\ &=\lim_{x\to0}\frac{2x\sec^2x\tan x+\sec^2x-\sec^2 x}{2x^2\sec^2x+4x\tan x}\\ &=\lim_{x\to0}\frac{2x\sec^2x\tan x}{2x^2\sec^2x+4x\tan x}\\ &=\lim_{x\to0}\frac{\sec^2x\tan x}{x\sec^2x+2\tan x}\end{aligned}$ And then we use L'Hospital's Rule again... What are you looking at? I know what I'm doing. Just bare with me. $\begin{aligned}&=\lim_{x\to0}\frac{\frac{d}{dx}\left(\sec^2x\tan x\right)}{\frac{d}{dx}\left(x\sec^2x+2\tan x\right)}\\ &=\lim_{x\to0}\frac{2\sec^2x\tan^2x+\sec^4x}{2x\sec^2x\tan x+\sec^2x+2\sec^2 x}\\ &=\lim_{x\to0}\frac{2\sec^2x\tan^2x+\sec^4x}{2x\sec^2x\tan x+3\sec^2x}\\ &=\lim_{x\to0}\frac{2\tan^2x+\sec^2x}{2x\tan x+3}\end{aligned}$ Now we can finally substitute in $x=0$ , knowing that $\tan 0=0$ and $\sec 0=1$ : $\begin{aligned}&=\frac{2(0)+(1)^2}{3+2(0)(0)}\\ &=\frac{1}{3}\end{aligned}$ This is not the answer yet, because remember we took the natural log of the question right at the start. So the final step is to reverse that: $\begin{aligned}\ln\left[\lim_{x\to0}\left(\frac{\tan x}{x}\right)^\frac{1}{x^2}\right]&=\frac{1}{3}\\ \implies\lim_{x\to0}\left(\frac{\tan x}{x}\right)^\frac{1}{x^2}&=\boxed{e^\frac{1}{3}}\end{aligned}$