How many ways are there to solve this?

$\lim_{x\to 0} \dfrac{\sin x}{\log(1+\tan x)} = \lim_{x\to 0} \dfrac{\dfrac{\sin x}{\tan x}}{\dfrac{\log(1+\tan x)}{\tan x}} = \dfrac{\lim_{x \to 0} \cos x}{\lim_{x\to 0} \dfrac{\log (1+\tan x)}{\tan x }} =\boxed{1}$

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Relevant wiki : L-Hopital's Rule $\lim_{x\to 0} \dfrac{\sin x}{\log(1+\tan x)} = \lim_{x\to 0} \dfrac{\cos x}{\dfrac{\sec ^2x}{1+\tan x}}=\lim_{x\to 0}\dfrac{\cos x\,(1+\tan x)}{\sec ^2x}=\boxed{1}$

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Relevant wiki : Taylor Series We have $\sin x = x -\dfrac{x^3}{3!} + \dfrac{x^5}{5!} -\dfrac{x^7}{7!} +\cdots \\ \text{\&}\quad \log(1 +\tan x) = \tan x-\dfrac{\tan ^2x}{2} +\dfrac{\tan ^3x}{3} -\dfrac{\tan ^4x}{4}+\cdots \\ \log(1+\tan x) = x+\dfrac{x^3}{3} +\dfrac{2x^5}{15} +\cdots +\dfrac{\left(x+\dfrac{x^3}{3} +\dfrac{2x^5}{15} +\cdots\right) ^2}{2} +$ Now $L = \lim_{x\to 0} \left[ \dfrac{x -\dfrac{x^3}{3!} + \dfrac{x^5}{5!} -\dfrac{x^7}{7!} +\cdots}{ x+\dfrac{x^3}{3} +\dfrac{2x^5}{15} +\cdots +\dfrac{\left(x+\dfrac{x^3}{3} +\dfrac{2x^5}{15} +\cdots\right) ^2}{2} +\cdots } \right]\\ L = \lim_{x\to 0}\dfrac{x}{x} \left[ \dfrac{1 -\dfrac{x^2}{3!} + \dfrac{x^4}{5!} -\dfrac{x^6}{7!} +\cdots}{ 1+\dfrac{x^2}{3} +\dfrac{2x^4}{15} +\cdots +\dfrac{ x\left(x+\dfrac{x^2}{3} +\dfrac{2x^4}{15} +\cdots\right) ^2}{2} +\cdots } \right] \\ L = \dfrac{1}{1} =\boxed{1}$

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At x=0, sin(x) and tan(x) are like x, and log(1+f(x)) is like f(x). So our limit is equivalent to the limit x/x, which is 1

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L & = \lim_{x \to 0} \frac {\sin x}{\log (1+\tan x)} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \frac {\cos x}{\frac {\sec^2 x}{1+\tan x}} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x. \\ & = \boxed{1}\end{aligned}$