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

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.

## closed as off-topic by Guy Fsone, Nosrati, Did, Micah, NamasteJan 27 '18 at 16:51

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• You should say "without L'Hopital" in the title and in the body of the question. – zhw. Feb 16 '17 at 17:21
• Do you want to allow power series? – Michael Burr Feb 16 '17 at 17:22
• Just use Taylor series – openspace Feb 16 '17 at 17:23
• You might TeX something from the picture. Not all people understand what the (Arabic) numbers mean. – Hopeless Feb 16 '17 at 17:23
• – lab bhattacharjee Feb 16 '17 at 17:44

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.

• no derivatives, "just" an outrageous overkill :P – dvb Jan 18 at 10:31

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.

• And how do you show it is legitimate? – MathematicsStudent1122 Feb 16 '17 at 23:35
• @MathematicsStudent1122 Give yourself an $\epsilon.$ Then $(1-\epsilon)s < \sin s < (1+\epsilon) s$ for small positive $s.$ Proceed ... – zhw. Feb 20 '17 at 19:21

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 $$\frac{1}{6}$$