Limits of trigonometric function I know my answer is correct, but are my steps correct?
$$
\begin{align}
& \lim_{t \to 0} \frac{\tan(2t)}{t}\\[8pt]
& = \lim_{t \to 0} \frac{1}{t} \tan(2t)\\[8pt]
& = \lim_{t \to 0} \frac{\sin(2t)}{t \cdot \cos(2t)} \cdot \frac{2}{2}\\[8pt]
& = \lim_{t \to 0} \frac{2 \cdot \sin(2t)}{2t \cdot \cos(2t)}\\[8pt]
& = \lim_{t \to 0} \left(\frac{\sin(2t)}{2t} \cdot \frac{2}{\cos(2t)}\right)\\[8pt]
& = 1 \cdot \frac{2}{1}\\[8pt]
& = 2
\end{align}
$$

Yes exactly is correct congratulation
