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**SlipEternal** To show $\displaystyle 1-\dfrac{\sin x}{x} \le \dfrac{x^2}{6}$ without using its power series expansion, we can use the Mean Value Theorem several times.

Let $\displaystyle f(x) = x^3-6x+6\sin x$. Then $\displaystyle f'(x) = 3x^2-6+6\cos x$, $\displaystyle f''(x) = 6x-6\sin x$, and $\displaystyle f'''(x) = 6-6\cos x$. Using a little trigonometry, we find $\displaystyle f'''(x) > 0$ for all $\displaystyle 0 < x < 2\pi$.

Choose any $\displaystyle 0 < x < 2\pi$. By the Mean Value Theorem, there exists $\displaystyle c_1\in (0,x)$ such that $\displaystyle f(x) - f(0) = (x-0)f'(c_1)$. Again, by the Mean Value Theorem, there exists $\displaystyle c_2 \in (0,c_1)$ such that $\displaystyle f'(c_1)-f'(0) = (c_1-0)f''(c_2)$. For the third time, by the Mean Value Theorem, there exists $\displaystyle c_3\in (0,c_2)$ such that $\displaystyle f''(c_2)-f''(0) = (c_2-0)f'''(c_3)$. Since $\displaystyle f(0)=f'(0)=f''(0)=0$, we have $\displaystyle f(x) = xf'(c_1) = xc_1f''(c_2) = xc_1c_2f'''(c_3)$. Since $\displaystyle 0<c_3<c_2<c_1<x$, we have the product of four positive numbers is positive, so $\displaystyle f(x)>0$ for any $\displaystyle 0<x<2\pi$.

This tells us $\displaystyle x^3-6x+6\sin x > 0 \Rightarrow x^3 > 6x-6\sin x \Rightarrow \dfrac{x^2}{6} > 1-\dfrac{\sin x}{x}$ for all $\displaystyle 0 < x < 2\pi$.