
   $$
      \lim_{x\to0}\frac{\sin 5x}x = \lim_{x\to0}\frac{5\sin5x}{5x}
      = 5\lim_{x\to0}\frac{\sin5x}{5x}
   $$
   $$
   \left.
   \begin{array}{rcl}
      \lim_{x\to0}5x             & = & 0 \\
      \lim_{y\to0}\frac{\sin y}y & = & 1 \\
      f(x) = 5x, \quad \forall x \in P(0) \colon\; x  & \neq & 0
   \end{array}
   \right\}
   \Longrightarrow 5\lim_{x\to0}\frac{\sin5x}{5x}=5.
   $$