Solve Trig Limit Problem: lim (x->0) \frac{x - sinx}{x^{3}}

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Homework Statement



is there a way to solve

lim (x->0) \frac{x - sinx}{x^{3}}


using the fact that sinx / x = 1

or is it much more complicated.. I've tried to break it down into sinx/x everyway i can think of with no luck..
 
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Do you know l'Hopitals rule?

I am thinking of a way to do it with only sin(x)/x but can't come up with a good way immediately.
 
I am not familiar with l'hopitals theorem yet...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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