Application of Taylor's theorem Use Taylor's theorem to prove the estimate
$${|\sin(x)-x+x^3/6|} \le C|x|^5$$
for a suitable constant C. Hence prove that $$\lim_{x \to 0} \frac{\sin(x)-x}{x^3}$$ exists and determine its value.
I used Taylor's theorem with remainder in Lagrange form and got $$\left |\sin(x)-x+\frac{x^3}{6} \right| \le \frac{1}{5!}\left|x\right |^5$$ 
and then I am not sure what I should do next in order to get the inequality into the required form .
I am also not sure how  I can use the inequality to find the limit as $x$ approaches zero.
Any help is appreciated, thanks in advance.
 A: do integration by parts quite a few times staring with 
$$\begin{align} \sin x  &= \int_0^x d \sin t = \int_0^x \cos t\, d(t-x)\\
&= (t-x)\cos t\big|_0^x + \int_0^x (t-x) \sin t \, dt\\
&=x + \frac 12 (t-x)^2 \sin t\big|_0^x - \frac 12 \int_0^x (t-x)^2 \cos t\, dt\\
&=x - \frac 16 (t-x)^3 \cos t\big|_0^x - \frac 16\int_0^x (t-x)^3 \sin t\, dt\\
&=x - \frac 16 x^3- \frac 1{24}\sin t (t-x)^4\big|_0^x +\frac 1{24}\int_0^x (t-x)^4 \cos t\, dt\\
&=x - \frac 16 x^3 +\frac 1{24}\int_0^x (t-x)^4 \cos t\, dt\\
\end{align}$$ 
  we can use the mean value theorem for integrals to find 
$$\int_0^x (t-x)^4 \cos t\, dt = \cos c \int_0^x (t-x)^4 \,dt = \frac 15 x^5 \cos c$$
putting all these together we get $$\big|\sin x - x + \frac 16 x^3\big| \le \frac{1}{120} |x|^5$$
A: From the Leibniz convergence test we know that the alternating series for $\sin x-x$ or better $\frac{\sin x - x}{x^3}$ converges as long as the absolute values $1/6, |x|^2/5!$, $|x|^4/7!$ etc. of its terms form a falling sequence converging to zero. Since that is true if it is true for the first two terms it is true for $x^2<20$.
Then, the test moreover tells us that the limit of the series is bracketed by the values of the partial sum, thus
$$
-\frac16\le\frac{\sin x-x}{x^3}\le-\frac16+\frac{x^2}{5!}
$$
or reordered
$$
0\le \frac{\sin x-x+\frac16 x^3}{x^5}\le\frac1{120}.
$$
