Which one of the following is correct?

1. $\displaystyle{} \lim_{x\rightarrow 0} \left( \dfrac{\sin4x}{\sin2x}\right)=2\;\text{and}\: \lim_{x\rightarrow 0} \left( \dfrac{\tan x}{x}\right)=1$
2. $\displaystyle{} \lim_{x\rightarrow 0} \left( \dfrac{\sin4x}{\sin2x}\right)=1\;\text{and}\; \lim_{x\rightarrow 0} \left( \dfrac{\tan x}{x}\right)=1$
3. $\displaystyle{} \lim_{x\rightarrow 0} \left( \dfrac{\sin4x}{\sin2x}\right)= \infty \;\text{and}\: \lim_{x\rightarrow 0} \left( \dfrac{\tan x}{x}\right)=1$
4. $\displaystyle{} \lim_{x\rightarrow 0} \left( \dfrac{\sin4x}{\sin2x}\right)=2\;\text{and}\; \lim_{x\rightarrow 0} \left( \dfrac{\tan x}{x}\right)= \infty$

$\\ \lim_{x\rightarrow 0} \dfrac{sin (4x)}{sin(2x)} \\ \\ \dfrac{0}{0}\ inderminant\ form\\ \\ Apply\ L'\ Hospital's\ rule\\ \\ \lim_{x\rightarrow0}\dfrac{4.cos(4x)}{2.cos(2x)}\\ \\ \dfrac{4}{2}=2$
$\\ \lim_{x\rightarrow 0} \dfrac{tan (x)}{x} \\ \\ \dfrac{0}{0}\ inderminant\ form\\ \\ Apply\ L'\ Hospital's\ rule\\ \\ \lim_{x\rightarrow0}\dfrac{sec^2x}{1}\\ \\ \dfrac{1}{1}=1$