Evaluate $displaystyleli{m}_{x\to 0}\frac{\sin (5x)-5x}{x^3}$ using L'Hôpital's rule.

Step-by-step solution

1. Recognize the indeterminate form $0/0$.
2. Apply L'Hôpital's rule once: numerator $5\cos (5x)-5$, denominator $3x^2$.
3. Limit becomes ${\lim }_{x\to 0}\frac{5(\cos 5x-1)}{3x^2}$, still $0/0$.
4. Apply L'Hôpital again: numerator $-25\sin 5x$, denominator $6x$.
5. Limit becomes ${\lim }_{x\to 0}\frac{-25\sin 5x}{6x}$, still $0/0$.
6. Apply L'Hôpital a third time: numerator $-125\cos 5x$, denominator $6$.
7. Evaluate at $x=0$: $\frac{-125\cos 0}{6}=-\frac{125}{6}$.

Concept tested: L'Hôpital's rule for higher-order limits

Published 3 September 2026 · How this solution was prepared and checked

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