Trigonometrical limit $\lim\limits_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}?$ Can you help me solve this without using de l'Hôpital's rule (just using Standard rules):
$$ \lim_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}? $$

Even though I like the approach of Dominic Michaelis, here is another one using infinite products.
First note that
\begin{align}
\sin(x) &= x \prod_{n=1}^\infty (1-\frac{x^2}{\pi^2n^2}) \\
&= x(1-\sum_{n=1}^\infty \frac{x^2}{\pi^2 n^2} + O(x^4))
\end{align}
and
\begin{align}
\cos(x) &=  \prod_{n=1}^\infty (1-\frac{x^2}{\pi^2(n-1/2)^2})
\\
\Leftrightarrow  x\cos(x) &=  x\prod_{n=1}^\infty (1-\frac{x^2}{\pi^2(n-1/2)^2})
\\ &= x(1-\sum_{n=1}^\infty \frac{x^2}{\pi^2 (n-1/2)^2} + O(x^4))
\end{align}
Therefore
\begin{align}
\frac{\sin(x)-x\cos(x)}{x^3}&= \frac{1}{x^2}\Bigl[\sum_{n=1}^\infty \frac{x^2}{\pi^2 (n-1/2)^2} - \sum_{n=1}^\infty \frac{x^2}{\pi^2 n^2} +O(x^4) \Bigr]
\\
&=\sum_{n=1}^\infty \frac{1}{\pi^2 (n-1/2)^2} - \sum_{n=1}^\infty \frac{1}{\pi^2 n^2} +O(x^2)  \\
&=\frac{1}{\pi^2} \Bigl[\sum_{n=1}^\infty \frac{1}{ (n-1/2)^2} - \sum_{n=1}^\infty \frac{1}{n^2} \Bigr]+O(x^2)
\end{align}
Together with the fact that 
\begin{align}
&\sum_{n=1}^\infty \frac{1}{ (n-1/2)^2} =\frac{\pi^2}{2} \quad \text{and} \\
&\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}
\end{align}
we get 
\begin{align}
\frac{\sin(x)-x\cos(x)}{x^3}
&=\frac{1}{\pi^2} \Bigl[\sum_{n=1}^\infty \frac{1}{ (n-1/2)^2} - \sum_{n=1}^\infty \frac{1}{n^2} \Bigr]+O(x^2) \\
&= \frac{1}{\pi^2} (\frac{\pi^2}{2}-\frac{\pi^2}{6}) +O(x^2)
\\ &= \frac{1}{3} + O(x^2) = 0 \quad \text{for } x\rightarrow 0
\end{align}
Note: I am aware, that this is not as simple as requested by Maths'problem and even more complex than the answer of Dominic Michaelis. We also need to know the infinite products and the identities involving $\sum 1/n^2$ and $\sum 1/(n-1/2)^2$. However, I just wanted to provide another method to solve the problem.
