Limit of a trig function. (Without using L'Hopital) I'm having trouble figuring out what to do here, I'm supposed to find this limit:
$$\lim_{x\rightarrow0} \frac{x\cos(x)-\sin(x)}{x^3}$$
But I don't know where to start, any hint would be appreciated, thanks!
 A: Taylor's formula at order $3$ will do (and anyway is better than L'Hospital's rule):
\begin{align}
\frac{x\cos x-\sin x}{x^3}&=\frac{1}{x^3}\biggl(x\Bigl(1-\frac{x^2}{2}+o(x^2)\Bigr)-\Bigl(x-\frac{x^3}{6}+o(x^3)\Bigr)\biggr)\\
&=\frac{1}{x^3}\biggl(\Bigl(\not x-\frac{x^3}{2}+o(x^3)\Bigr)-\Bigl(\not x-\frac{x^3}{6}+o(x^3)\Bigr)\biggr)\\
&=\frac{1}{x^3}\Bigl(-\frac{x^3}{3}+o(x^3)\Bigr)=-\frac13+o(1)\to -\frac13.
\end{align}
A: Since$$x\cos(x)-\sin(x)=x\left(1-\frac{x^2}2+\cdots\right)-\left(x-\frac{x^3}6+\cdots\right)=-\frac{x^3}3+\cdots,$$your limit is equal to $-\dfrac13$.
A: One can use L'Hospital's rule or, as shown here, the series expansion of $\sin(x)$ and $\cos(x)$. 
Using 
\begin{align}
\cos(x) &= 1 - \frac{x^{2}}{2} + \frac{x^{4}}{4!} - \cdots \\
\sin(x) &= x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots
\end{align}
then
$$x \cos(x) - \sin(x) = - \frac{2 \, x^{3}}{3!} + \frac{4 \, x^{5}}{5!} - \cdots$$
and
$$\lim_{x \to 0} \frac{x \cos(x) - \sin(x)}{x^{3}} = \lim_{x \to 0} \left( - \frac{2}{3!} + \frac{4 \, x^{2}}{5!} - \cdots \right) = - \frac{1}{3}.$$
A: You could do 
Let $u=x^3$
So this gives us 
$\lim_{u \rightarrow 0} \frac{ \sqrt[3]{u} \cos(\sqrt[3]{u})-\sin(\sqrt[3]{u})}{
u}$
Let $f(u)=\sqrt[3]{u} \cos(\sqrt[3]{u})-\sin(\sqrt[3]{u})$
And so $f(0)=0$
We know that 
$\lim_{u \rightarrow 0} \frac{ f(u)-f(0)}{u-0}=\frac{d}{du}(f(u))|_{u=0}$
A: Hint:
In Equation $(10)$ of this answer, it is shown that
$$
\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag1
$$
Furthermore, as shown in this answer, $\lim\limits_{x\to0}\frac{\sin(x)}x=1$, so
$$
\begin{align}
\lim_{x\to0}\frac{1-\cos(x)}{x^2}
&=\lim_{x\to0}\frac{2\sin^2(x/2)}{4(x/2)^2}\\
&=\frac12\left(\lim_{x\to0}\frac{\sin(x/2)}{x/2}\right)^2\\[3pt]
&=\frac12\tag2
\end{align}
$$
Limits $(1)$ and $(2)$ can be combined to answer the question.
