Step 1: Look at the problem.
We must find $\lim_{x\to0}\dfrac{\sin5x-5\sin x}{x^3}$. If we just plug in $x=0$ we get $\tfrac{0}{0}$, which is undefined, so we need a smarter method.
Step 2: Pick a clean method.
Because both top and bottom go to zero, we may use L'Hopital's rule, which lets us differentiate the top and the bottom separately until the answer appears.
Step 3: First differentiation.
Differentiate top and bottom once: \[ \frac{5\cos5x-5\cos x}{3x^2}. \] Putting $x=0$ still gives $\tfrac{5-5}{0}=\tfrac{0}{0}$, so we differentiate again.
Step 4: Second differentiation.
\[ \frac{-25\sin5x+5\sin x}{6x}. \] At $x=0$ this is again $\tfrac{0}{0}$, so we go one more time.
Step 5: Third differentiation.
\[ \frac{-125\cos5x+5\cos x}{6}. \] Now we can safely put $x=0$.
Step 6: Put in the value.
At $x=0$, $\cos5x=1$ and $\cos x=1$, so the top is $-125+5=-120$. Then \[ \frac{-120}{6}=-20. \] So the limit equals $-20$.
\[ \boxed{-20} \]