Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{\operatorname{atan}{\left(3 x \right)}}{15 x}\right)$$
Sustituimos
$$u = \operatorname{atan}{\left(3 x \right)}$$
$$x = \frac{\tan{\left(u \right)}}{3}$$
obtendremos
$$\lim_{x \to 0^+}\left(\frac{\operatorname{atan}{\left(3 x \right)}}{15 x}\right) = \frac{\lim_{u \to 0^+}\left(\frac{\operatorname{atan}{\left(\frac{3 \tan{\left(u \right)}}{3} \right)}}{\frac{1}{3} \tan{\left(u \right)}}\right)}{15}$$
=
$$\frac{\lim_{u \to 0^+}\left(\frac{3 \operatorname{atan}{\left(\tan{\left(u \right)} \right)}}{\tan{\left(u \right)}}\right)}{15} = \frac{\lim_{u \to 0^+}\left(\frac{3 u}{\tan{\left(u \right)}}\right)}{15}$$
=
$$\frac{\lim_{u \to 0^+} \frac{1}{\frac{1}{u} \tan{\left(u \right)}}}{5}$$
/tan(u)\
= 1/5 / ( lim |------| )
u->0+\ u /
cambiamos
$$\lim_{u \to 0^+}\left(\frac{\tan{\left(u \right)}}{u}\right) = \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u \cos{\left(u \right)}}\right)$$
=
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right) \lim_{u \to 0^+} \cos{\left(u \right)} = \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
El límite
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
hay el primer límite, es igual a 1.
Entonces la respuesta definitiva es:
$$\lim_{x \to 0^+}\left(\frac{\operatorname{atan}{\left(3 x \right)}}{15 x}\right) = \frac{1}{5}$$