Answer
$\dfrac{-1}{6} $
Work Step by Step
Consider: $\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\dfrac{\sin x-x}{x^3}$
Need to check that the limit has an indeterminate form.
Thus, $f(0)=\dfrac{\sin (0)-0}{0^3}=\dfrac{0}{0}$
Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
This implies:
$\lim\limits_{x \to 0}\dfrac{\cos x -1}{3x^2}=\dfrac{0}{0} $
Again apply L-Hospital's rule:
we get $\lim\limits_{x \to 0}\dfrac{-\sin x}{6x}=\dfrac{-\sin 0}{6(0)}=\dfrac{0}{0} $
Now, again apply L-Hospital's rule .
we get $\lim\limits_{x \to 0}\dfrac{-\cos x}{6}=\dfrac{-\cos 0}{6}=\dfrac{-1}{6} $
