Calculating the Limit: $\lim_{x\to0}\frac{\sin x-x }{x^3}$

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Hi


Find the limit:
\lim_{x\to0}\frac{\sin x-x }{x^3}
 
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Mohammad_93 said:
Hi


Find the limit:
\lim_{x\to0}\frac{\sin x-x }{x^3}
What have you tried?
 
lim[x->0] (sinx - x)/x^3 = lim [x->0] sinx/x3 - 1/x2
but it's infinity - infinity!
 
So that didn't do you any good.

Do you know about L'Hopital's Rule?
 
Or the power series for sin x?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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