Answer
$$\lim_{t\to0}\frac{2t}{\tan t}=2$$
Work Step by Step
$$A=\lim_{t\to0}\frac{2t}{\tan t}=\lim_{t\to0}\Big(\frac{\tan t}{2t}\Big)^{-1}=\Big(\lim_{t\to0}\frac{\tan t}{2t}\Big)^{-1}=X^{-1}$$
Considering $X$:
$$X=\lim_{t\to0}\frac{\tan t}{2t}$$ $$X=\lim_{t\to0}\frac{\frac{\sin t}{\cos t}}{2t}=\lim_{t\to0}\frac{\sin t}{2t\cos t}$$ $$X=\lim_{t\to0}\frac{\sin t}{2t}\lim_{t\to0}\frac{1}{\cos t}$$ $$X=\lim_{t\to0}\frac{\sin t}{2t}\times\Big(\frac{1}{\cos0}\Big)$$ $$X=\lim_{t\to0}\frac{\sin t}{2t}\times\Big(\frac{1}{1}\Big)$$ $$X=\lim_{t\to0}\frac{\sin t}{2t}$$
To be able to apply Theorem 7, we need $t$ in the denominator, instead of $2t$.
So we would multiply both numerator and denominator by $1/2$:
$$X=\lim_{t\to0}\frac{1/2\sin t}{1/2\times2t}=\frac{1}{2}\lim_{t\to0}\frac{\sin t}{t}$$
Apply Theorem 7: $$X=\frac{1}{2}\times1=\frac{1}{2}$$
Therefore, $$A=X^{-1}=\Big(\frac{1}{2}\Big)^{-1}=2$$
