## Calculus 10th Edition

$\lim\limits_{x\to0^+}\dfrac{\sin{4x}}{5x}=\dfrac{4}{5}.$
Can be solved in two ways: $I:$ $\lim\limits_{x\to0^+}\dfrac{\sin{4x}}{5x}=\frac{4}{5}(\lim\limits_{x\to0^+}\dfrac{\sin{4x}}{4x})\to$ let $z=4x\to$ as $x$ approaches $0^+$ so does $z\to$ $\frac{4}{5}(\lim\limits_{z\to0^+}\dfrac{\sin{z}}{z})=\frac{4}{5}(1)=\dfrac{4}{5}.$ $II:$ $\lim\limits_{x\to0^+}\dfrac{\sin{4x}}{5x}=\dfrac{\sin{[4(0^+)]}}{5(0^+)}=\dfrac{4(0^+)}{5(0^+)}=\dfrac{4}{5}.$ Note: for $\theta$ (in radians) very close to 0, $\sin{\theta}\approx\theta.$ Since $0^+$ is extremely close to $0\to \sin{[c(0^+)]}=c(0^+)$ where $c$ is a constant.