Evaluating trigonometric limits with variable exponent $\left(\frac{\tan x}{x}\right)^\frac1x$ I found a set of problems of limits which i can't seem to work my way around.
I tried using the natural log and then applying L'Hospital's rule but I can't seem to make it work. 
The problem was to find the limit of the function of 
   $$\left(\frac{\tan x}{x}\right)^\frac1x$$
As x approaches 0
Please help me as there are some more problems like this. I cannot think of ways to evaluate them
 A: L'Hospital's rule is not the alpha and omega of limits computations!
To determine the limit of the logarithm,  use Taylor's formula at order $3$ for the tangent:
$$\frac{\tan x}x=\frac{x+\cfrac{x^3}3+o(x^3)}x=1+\frac{x^2}3+o(x^2),$$
so that
$$\frac1x\log\Bigl(\frac{\tan x}x\Bigr)=\frac1x\log\Bigl(1+\frac{x^2}3+o(x^2)\Bigr)=\frac1x\Bigl(\frac{x^2}3+o(x^2)\Bigr)=\frac{x}3+o(x)\to 0$$
and finally $\;\biggl(\dfrac{\tan x}x\biggr)^{\!\tfrac 1x}$ tends to $1$ as $x$ tends to $0$.
A: We have that (refer to this OP)
$$\lim_{x\to0}\frac{\tan x-x}{x^3}=\frac13 \implies \tan x=x+ \frac13x^3+o(x^3)$$
therefore
$$\left(\frac{\tan x}{x}\right)^\frac1x=\left(1+ \frac13x^2+o(x^2)\right)^\frac1x=\left[\left(1+ \frac13x^2+o(x^2)\right)^{\frac1{\frac13x^2+o(x^2)}}\right]^{\frac{\frac13x^2+o(x^2)}x}\to e^0=1$$
Frome the same derivation we can now easily derive that
$$\left(\frac{\tan x}{x}\right)^\frac1{x^2}\to \sqrt[3] e$$
A: Recall that $$\lim_{x \to \infty} \left( 1 + \frac{k}{x}\right)^x = \lim_{x \to 0^+} \left(1 + kx\right)^{1/x} = e^k$$ and for any $\epsilon > 0$ there is some $\delta > 0$ such that $$0 < x < \delta \implies x < \tan x < x + \epsilon x^2.$$ Therefore, $$1 = \lim_{x \to 0^+} 1^x \leq \lim_{x \to 0^+} \left( \frac{\tan x}{x} \right)^{1/x} \leq \lim_{x \to 0^+} (1 + \epsilon x)^{1/x} = e^\epsilon.$$
Now take $\epsilon \to 0$.
A: The limit of logarithm can be found without Taylor's formula: 
$$
\begin{align*}
\lim_{x\to 0}\frac{\log\left(\frac{\tan x}{x}\right)}{x}&=
\underbrace{\lim_{x\to 0}\frac{\log\left(\frac{\tan x}{x}\right)}{\frac{\tan x}{x}-1}}_{=1}\cdot\lim_{x\to 0}\frac{\frac{\tan x}{x}-1}{x}
=\lim_{x\to 0}\frac{\frac{\tan x}{x}-1}{x}
=\lim_{x\to 0}\frac{\tan x-x}{x^2}
\stackrel{\text{L'Hosp}}{=}\lim_{x\to 0}\frac{\frac{1}{\cos^2x}-1}{2x}\\[12pt]
&=\lim_{x\to 0}\frac{1}{\cos^2x}\cdot\lim_{x\to 0}\frac{1-\cos^2x}{2x}
=\lim_{x\to 0}\frac{\sin^2x}{2x}
=\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{\sin x}{2}=0
\end{align*}
$$
The second limit is derived from the identity
$$
\lim_{t\to 1}\frac{\log t}{t-1}=1
$$
after substituion $t=\frac{\tan x}{x}$ as $x\to 0$.
