No LIMITations!

Here are the 3 ways that I found of doing this. I would love to see other ways posted as well if anyone knows any. Make them as bizarre as possible!

Method 1: This is a traditional way of solving this limit. It involves rewriting tangent in terms of sine and cosine.

We know that $\tan(x) \equiv \cfrac{\sin(x)}{\cos(x)}$ . We can use this to our advantage: $\begin{aligned} \lim\limits_{x\to 0}\frac{\sin(x)}{\tan(x)} \Longrightarrow \lim\limits_{x\to 0}\sin(x)\frac{1}{\frac{\sin(x)}{\cos(x)}} &= \lim\limits_{x\to 0}\sout{\sin(x)}\frac{\cos(x)}{\sout{\sin(x)}} \\ &= \lim\limits_{x\to 0}\cos(x) = \boxed{1} \end{aligned}$

Method 2: This is the method that I look to first; I guess it is out of habit.

We can use L'Hopital's Rule here because $\sin(0) = 0$ and $\tan(0) = 0$ making this a $\cfrac{0}{0}$ situation; we can differentiate the top and bottom and work from there: $\begin{aligned} \lim\limits_{x\to 0}\frac{\sin(x)}{\tan(x)} \Longrightarrow \lim\limits_{x\to 0}\frac{\frac{d}{dx}(\sin(x))}{\frac{d}{dx}(\tan(x))} &= \lim\limits_{x\to 0}\frac{\cos(x)}{\sec^2(x)} \\ &= \lim\limits_{x\to 0}\frac{\cos(x)}{\frac{1}{\cos^2(x)}} \;\;\; \left(\sec(x) \equiv \frac{1}{\cos(x)}\right) \\ &= \lim\limits_{x\to 0}\cos^3(x) = 1^3 = \boxed{1} \end{aligned}$

Method 3: This method is similar to method 1, but it is more bizarre which is what I would like! This uses the imaginary expressions of sine, cosine and tangent.

First, we must know that $\sin(x) \equiv \cfrac{e^{ix}-e^{-ix}}{2i}$ , $\cos(x) \equiv \cfrac{e^{ix}+e^{-ix}}{2}$ and $\tan(x) \equiv \cfrac{i(e^{-ix}-e^{ix})}{e^{ix}+e^{-ix}}$ where $i = \sqrt{-1}$ . Before we plug these into our limit, let us also know that $\cfrac{i(e^{-ix}-e^{ix})}{e^{ix}+e^{-ix}} = \cfrac{-i(e^{ix}-e^{-ix})}{e^{ix}+e^{-ix}}$ because this will help us later.

Now, we can plug them in: $\begin{aligned} \lim\limits_{x\to 0}\frac{\sin(x)}{\tan(x)} \Longrightarrow \lim\limits_{x\to 0}\frac{e^{ix}-e^{-ix}}{2i} \frac{1}{\frac{i(e^{-ix}-e^{ix})}{e^{ix}+e^{-ix}}} &= \lim\limits_{x\to 0}\frac{e^{ix}-e^{-ix}}{2i} \frac{e^{ix}+e^{-ix}}{i(e^{-ix}-e^{ix})} \\ &= \lim\limits_{x\to 0}\frac{e^{ix}-e^{-ix}}{2i} \frac{e^{ix}+e^{-ix}}{i(e^{-ix}-e^{ix})} \\ &= \lim\limits_{x\to 0}\frac{\sout{e^{ix}-e^{-ix}}}{2i} \frac{e^{ix}+e^{-ix}}{-i\sout{(e^{ix}-e^{-ix})}} \\ &= \lim\limits_{x\to 0}\frac{e^{ix}+e^{-ix}}{2} \\ &= \lim\limits_{x\to 0}\cos(x) = \boxed{1} \end{aligned}$

Again, if there are any more crazy ways of solving this, then please post them as a solution!

Draw a circle of radius $1$ with center at $O$ . Take two points $A$ and $B$ on it's circumference. Join $\overline {OA}$ and extend it to $C$ . Draw a tangent to the circle at $B$ meeting $\overline {OC}$ at $D$ . Join $\overline {OB},\overline {AB}$ .

Area of $\triangle {OAB}$ is $\dfrac 12 \sin x$ , where

$x=\angle {AOB}$ . Area of circular sector $AOB$ is $\dfrac 12 x$ ,

and area of $\triangle {OBD}$ is $\dfrac 12 \tan x$

It's easy to compare these three areas to obtain $\sin x<x<\tan x$ for $0<x<\dfrac π2$

By moving the point $B$ towards the point $A$ along the arc of the circle, we see that as $x$ decreases, the point $D$ also moves towards $A$ , these three areas become more and more close, so that when the angle is infinitesimally small, they merge totally, so that the ratio of any two of them becomes $1$ :

$\displaystyle \lim_{x\to 0} \dfrac {\sin x}{x}$

$=\displaystyle \lim_{x\to 0} \dfrac {\tan x}{x}$

$=\displaystyle \lim_{x\to 0} \dfrac {\sin x}{\tan x}=\boxed 1$ .