Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{\operatorname{atan}{\left(\frac{5 x}{2} \right)}}{6 x}\right)$$
Sustituimos
$$u = \operatorname{atan}{\left(\frac{5 x}{2} \right)}$$
$$x = \frac{2 \tan{\left(u \right)}}{5}$$
obtendremos
$$\lim_{x \to 0^+}\left(\frac{\operatorname{atan}{\left(\frac{5 x}{2} \right)}}{6 x}\right) = \frac{\lim_{u \to 0^+}\left(\frac{\operatorname{atan}{\left(\frac{\frac{5}{2} \tan{\left(u \right)}}{\frac{5}{2}} \right)}}{\frac{1}{\frac{5}{2}} \tan{\left(u \right)}}\right)}{6}$$
=
$$\frac{\lim_{u \to 0^+}\left(\frac{\frac{5}{2} \operatorname{atan}{\left(\tan{\left(u \right)} \right)}}{\tan{\left(u \right)}}\right)}{6} = \frac{\lim_{u \to 0^+}\left(\frac{5 u}{2 \tan{\left(u \right)}}\right)}{6}$$
=
$$\frac{5 \lim_{u \to 0^+} \frac{1}{\frac{1}{u} \tan{\left(u \right)}}}{12}$$
/tan(u)\
= 5/12 / ( 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:
$$\lim_{x \to 0^+}\left(\frac{\operatorname{atan}{\left(\frac{5 x}{2} \right)}}{6 x}\right) = \frac{5}{12}$$