Answer
The solution is
$$\lim_{x\to0}\frac{\sin x-x}{7x^3}=-\frac{1}{42}.$$
Work Step by Step
We will calculate this limit using L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule".
$$\lim_{x\to0}\frac{\sin x-x}{7x^3}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(\sin x-x)'}{(7x^3)'}=\lim_{x\to0}\frac{\cos x-1}{21x^2}=\left[\frac{\cos 0-1}{21\cdot0^2}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(\cos x-1)'}{(21x^2)'}=\lim_{x\to0}\frac{-\sin x}{42x}=\left[\frac{-\sin 0}{42\cdot0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(-\sin x)'}{(42x)'}=\lim_{x\to0}\frac{-\cos x}{42}=\frac{-\cos 0}{42}=-\frac{1}{42}.$$
