Answer
$$\lim_{x\to0}\frac{\tan 2x}{x}=2$$
Work Step by Step
$$A=\lim_{x\to0}\frac{\tan 2x}{x}$$ $$A=\lim_{x\to0}\frac{\frac{\sin 2x}{\cos2x}}{x}=\lim_{x\to0}\frac{\sin 2x}{x\cos2x}$$ $$A=\lim_{x\to0}\frac{\sin2x}{x}\lim_{x\to0}\frac{1}{\cos2x}$$ $$A=\lim_{x\to0}\frac{\sin2x}{x}\times\Big(\frac{1}{\cos0}\Big)$$ $$A=\lim_{x\to0}\frac{\sin2x}{x}\times\Big(\frac{1}{1}\Big)$$ $$A=\lim_{x\to0}\frac{\sin2x}{x}$$
To be able to apply Theorem 7, we need $2x$ in the denominator, instead of $x$.
So we would multiply both numerator and denominator by $2$:
$$A=\lim_{x\to0}\frac{2\sin2x}{2\times x}=2\lim_{x\to0}\frac{\sin 2x}{2x}$$
Take $\theta=2x$. Then as $x\to0$, $\theta\to2\times 0=0$
$$A=2\lim_{\theta\to0}\frac{\sin\theta}{\theta}$$
Apply Theorem 7: $$A=2\times1=2$$
