Which one of the following is correct?

1. $\lim_{x\rightarrow 0} ( \frac{sin4x}{sin2x})=2$ and $\lim_{x\rightarrow 0} ( \frac{tanx}{x})=1$
2. $\lim_{x\rightarrow 0} ( \frac{sin4x}{sin2x})=1$ and $\lim_{x\rightarrow 0} ( \frac{tanx}{x})=1$
3. $\lim_{x\rightarrow 0} ( \frac{sin4x}{sin2x})=$ $\infty$ and $\lim_{x\rightarrow 0} ( \frac{tanx}{x})=1$
4. $\lim_{x\rightarrow 0} ( \frac{sin4x}{sin2x})=2$ and $\lim_{x\rightarrow 0} ( \frac{tanx}{x})= \infty$
in Calculus
retagged

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