We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(\left(e^{x} - 1\right) e^{3 x}\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+}\left(- \sin{\left(3 x \right)} + \sin{\left(4 x \right)}\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{e^{4 x} - e^{3 x}}{- \sin{\left(3 x \right)} + \sin{\left(4 x \right)}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{\left(e^{x} - 1\right) e^{3 x}}{- \sin{\left(3 x \right)} + \sin{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(e^{x} - 1\right) e^{3 x}}{\frac{d}{d x} \left(- \sin{\left(3 x \right)} + \sin{\left(4 x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{4 e^{4 x} - 3 e^{3 x}}{- 3 \cos{\left(3 x \right)} + 4 \cos{\left(4 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{4 e^{4 x} - 3 e^{3 x}}{- 3 \cos{\left(3 x \right)} + 4 \cos{\left(4 x \right)}}\right)$$
=
$$1$$
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)