Limit with De L'Hôpital

Statement

Using De L’Hôpital’s rule: $\lim_{x\to 0}\dfrac{x-\sin x}{x^3}$.

Solution

It is a $\frac{0}{0}$ form. Applying De L’Hôpital repeatedly (always a $\frac00$ form): $\lim_{x\to 0}\frac{x-\sin x}{x^3} = \lim_{x\to 0}\frac{1-\cos x}{3x^2} = \lim_{x\to 0}\frac{\sin x}{6x} = \lim_{x\to 0}\frac{\cos x}{6} = \boxed{\frac{1}{6}}.$ The same result is obtained in a single step with the Taylor expansion $\sin x = x - x^3/6 + o(x^4)$.

Links

Topics: Derivatives
Concepts: Limit · De L’Hôpital’s rule
Skills: Computing limits
People: Guillaume de l’Hôpital
Exercise type: Limit computation