Notable limit of the sine

Text

Compute $\displaystyle\lim_{x\to 0}\frac{\sin(5x)}{3x}$.

Solution

Use the notable limit $\displaystyle\lim_{t\to 0}\frac{\sin t}{t}=1$. Rewrite so the argument $5x$ appears in both the sine and the denominator: $\frac{\sin(5x)}{3x}=\frac{\sin(5x)}{5x}\cdot\frac{5x}{3x}=\frac{\sin(5x)}{5x}\cdot\frac{5}{3}.$ As $x\to 0$ the first factor tends to $1$, so $L=1\cdot\frac53=\frac53\approx 1.667.$ $\boxed{\,L=\tfrac53\,}$