I'm unsure as to how to evaluate:
$$\lim\limits_{x\to 0} \frac{\sin x – x + \frac{x^3}{6}}{x^3}$$
The $\lim\limits_{x\to 0}$ of both the numerator and denominator equal $0$. Taking the derivative of both ends of the fraction we get:
$$\lim\limits_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2}$$
But I don't know how to evaluate this?
Many thanks for any help.
Best Answer
Using L'Hospital twice, $$ \lim_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}=\lim_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2}=\lim_{x\to 0} \frac{x - \sin x }{6x}=\frac{1}{6}\lim_{x\to 0} \left(1-\frac{\sin x}{x}\right)=0 $$