Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\operatorname{atan}{\left(5 x \right)}}{3 x}\right)$$
Do replacement
$$u = \operatorname{atan}{\left(5 x \right)}$$
$$x = \frac{\tan{\left(u \right)}}{5}$$
we get
$$\lim_{x \to 0^+}\left(\frac{\operatorname{atan}{\left(5 x \right)}}{3 x}\right) = \frac{\lim_{u \to 0^+}\left(\frac{\operatorname{atan}{\left(\frac{5 \tan{\left(u \right)}}{5} \right)}}{\frac{1}{5} \tan{\left(u \right)}}\right)}{3}$$
=
$$\frac{\lim_{u \to 0^+}\left(\frac{5 \operatorname{atan}{\left(\tan{\left(u \right)} \right)}}{\tan{\left(u \right)}}\right)}{3} = \frac{\lim_{u \to 0^+}\left(\frac{5 u}{\tan{\left(u \right)}}\right)}{3}$$
=
$$\frac{5 \lim_{u \to 0^+} \frac{1}{\frac{1}{u} \tan{\left(u \right)}}}{3}$$
/tan(u)\
= 5/3 / ( lim |------| )
u->0+\ u / transform
$$\lim_{u \to 0^+}\left(\frac{\tan{\left(u \right)}}{u}\right) = \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u \cos{\left(u \right)}}\right)$$
=
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{u \to 0^+} \cos{\left(u \right)} = \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
The limit
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
is first remarkable limit, is equal to 1.
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\operatorname{atan}{\left(5 x \right)}}{3 x}\right) = \frac{5}{3}$$