An easy limit?

By using L'Hopital's Rule:

$\lim_{x \rightarrow 0} \frac{7x^3}{x-\sin x} = \lim_{x \rightarrow 0} \frac{21x^2}{1-\cos x} = \lim_{x \rightarrow 0} \frac{42x}{\sin x} = \lim_{x \rightarrow 0} \frac{42}{\cos x} = 42$

Series expansion of sine function near $x = 0$ : $\sin x = x - \frac{1}{6}x^3 + \cdots + \frac{(-)^n}{(2n+1)!}x^{2n+1}+\cdots$ Thus, $\lim_{x\to0}\frac{7x^3}{x-\sin x}=\lim_{x\to0}\frac{7x^3}{\frac{1}{6}x^3+O(x^5)}=7\cdot 6 = 42.$