$\lim\limits_{x\to 0}\dfrac{\sin x -x }{x^3}$ limiti

Question

$$\lim\limits_{x\to 0}\frac{\sin  x -x }{x^3}$$ limitini bulunuz.

Answer 1

Üç kere l'Hôpital kuralını kullanırsak\begin{align*}\lim\limits_{x\to 0}\frac{\sin  x -x }{x^3} \ &\mathop{=}_{\text{l'H}}^{\left[\frac00\right]} \ \lim\limits_{x\to 0}\frac{\cos x -1 }{3x^2}\\[17pt]&\mathop{=}_{\text{l'H}}^{\left[\frac00\right]} \ \lim\limits_{x\to 0}\frac{-\sin x}{6x}\\[17pt]&\mathop{=}_{\text{l'H}}^{\left[\frac00\right]}\ \lim\limits_{x\to 0}\frac{-\cos x}{6}\\[17pt]&= \ \frac{-\cos 0}6\\[17pt]&= \ -\frac16\end{align*}eşitliğini elde ederiz.