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Another application of the same idea

\begin{displaymath} \lim_{x\rightarrow 0}\frac{\sin(2x)}{x} = \lim_{x\rightarrow 0}2\frac{\sin(2x)}{2x} = 2\times 1\end{displaymath}

since we know that

\begin{displaymath} \lim_{u\rightarrow 0}\frac{\sin(u)}{u} = 1\end{displaymath}

and $\lim_{x\rightarrow 0}2x = 0$.
David G Radcliffe
8/18/1998