# How to evaluate $\lim\limits_{x\to 0} \frac{\sin x - x + x^3/6}{x^3}$

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.

• The limit is more interesting with $x^5$ at the denominator. – Yves Daoust Mar 11 '17 at 20:48

You can use l'Hospital as many times as needed as long as the indeterminate forms conditions are fulfilled. In this case, using Taylor series can be helpful, too:

$$\sin x = x - \frac{x^3}6 + \frac{x^5}{120} - \ldots = x - \frac{x^3}6 + \mathcal O(x^5)$$

$$\implies \frac{\sin x - x + \frac{x^3}6}{x^3} = \frac{\mathcal O(x^5)}{x^3} = \mathcal O(x^2) \xrightarrow[x \to 0]{} 0$$

Using L'Hospital twice, $$\lim_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}=\lim_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2}=\lim_{x\to 0} \frac{x - \sin x }{6x}=\frac{1}{6}\lim_{x\to 0} \left(1-\frac{\sin x}{x}\right)=0$$

Hint: It still has indeterminate form $\frac{0}{0}$, so we may apply L'Hopital's rule again. The result will still have indeterminate form $\frac{0}{0}$, so we may apply L'Hopital's rule yet again. You will finally get a result whose limit can be easily evaluated.

Using the identity $sin \ x =3sin(\frac{x}{3})-4 sin^{3}(\frac{x}{3})$, our limit $l=lim \frac{sin \ x -x +\frac{x^{3}}{6}}{x^{3}}$ becomes, $l= \frac{1}{9} lim \frac{sin \ \frac{x}{3} -\frac{x}{3} +(\frac{x}{3})^{3}.\frac{1}{6}}{x^{3}}+ \frac{4}{27}- lim \ 4 \frac{sin^{3}(\frac{x}{3})}{x^{3}}$, which means $l=\frac{1}{9} l$, so $l=0$.

You can continue, using $\cos x = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = 1-2\sin^2(\frac{x}{2}) \iff \cos x -1 = -2\sin^2(\frac{x}{2})$ after your last step so that $$\lim\limits_{x\to 0} \frac{x^2 + 2\cos x -2}{6x^2} = \lim_{x\to0} \frac{1+2\Big(\frac{\cos x - 1}{x^2}\Big)}{6} = \lim_{x\to0} \frac{1+2\Big(\frac{-2\sin^2(\frac{x}{2})}{x^2}\Big)}{6} = \lim_{t\to0} \frac{1-\Big(\frac{\sin t}{t}\Big)^2}{6} = \lim_{t\to 0} \frac{1-1}{6} = \color{red}{\fbox{0}}$$

where $t=\frac{x}{2}$