Find the limit: $\lim_{ x \to0 }\frac{x-\sin x}{x^3}$ without using the L'Hopital.

Question

Find the limit without using L'Hopital's rule

$$\lim_{ x \to0 }\frac{x-\sin x}{x^3}=?$$ This was what I did but I would like another solution.

Answer 1

If you like an outrageous overkill, from the Weierstrass product $$ \frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right)\tag{1} $$ it follows that in a neighbourhood of the origin we have: $$ 1-\frac{\zeta(2)}{\pi^2}x^2 \leq \frac{\sin x}{x}\leq \exp\left(-\frac{\zeta(2)}{\pi^2}x^2\right)\tag{2} $$ and the given limit equals $\frac{\zeta(2)}{\pi^2}=\color{red}{\large\frac{1}{6}}$ by squeezing.

Answer 2

No Taylor, no L'Hopital: Note that $x-\sin x = \int_0^x (1-\cos t)\, dt.$ Integrating again gives

$$x-\sin x = \int_0^x \int_0^t\sin s \, ds\, dt.$$

We know for small $s$ that $\sin s \sim s.$ So let's use $s$ in place of $\sin s$ above to see what's going on. We get

$$\int_0^x \int_0^t s \, ds\, dt = \int_0^x (t^2/2) dt = x^3/6.$$

Dividing that by $x^3$ gives a limit of $1/6.$ That shows nicely where the answer comes from. All that's left is to make sure using $s$ in place of $\sin s$ above is legitimate.

Answer 3

power series at $$x=0$$ of $$\frac{x-\sin(x)}{x^3}$$ gives $$\frac{1}{6}-\frac{x^2}{120}+\frac{x^4}{540}$$+Terms of the order $$x^6$$ thus our Limit is $$\frac{1}{6}$$