We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(\sin{\left(x \right)} + \sin{\left(3 x \right)}\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+}\left(10 x\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(x \right)} + \sin{\left(3 x \right)}}{10 x}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(x \right)} + \sin{\left(3 x \right)}}{10 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(\sin{\left(x \right)} + \sin{\left(3 x \right)}\right)}{\frac{d}{d x} 10 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)}}{10} + \frac{3 \cos{\left(3 x \right)}}{10}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)}}{10} + \frac{3 \cos{\left(3 x \right)}}{10}\right)$$
=
$$\frac{2}{5}$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)