How can I evaluate this limit without using L'hospital, derivatives or series expansion? I have troubles solving this particular limit without using series expansion, L'Hospital rule or derivatives.
$$\lim_{x \to 0}{\frac{x\sin{x}+2(\cos{x}-1)}{x^3}}$$
I have tried multiplying by $(\cos{x}+1)$ and got rid of one $x$ in the denominator, but got arguably something even messier. I got this:
$$\lim_{x \to 0}{\frac{x(\cos{x}+1)-2\sin{x}}{x^2(\cos{x}+1)}}$$
Any help would be appreciated.
 A: Various limits are proved in this answer using basic tools only. In particular, we need $\lim_{t\to 0}\sin(t)/t=0$, $\lim_{t\to 0}(\cos(t)-1)/t=0$, and $\lim_{t\to 0}(t-\sin(t))/t^3=1/6$.
$$\begin{align}
\frac{x\sin x+2(\cos x-1)}{x^3}&=\frac{x(2\sin(x/2)\cos(x/2))+2(-2\sin^2(x/2))}{x^3} \\
&= \frac{\sin(x/2)}{x/2}\cdot\frac{x\cos(x/2)-2\sin(x/2)}{x^2}
\end{align}$$
We have $\lim_{x\to 0}\sin(x/2)/(x/2)=1$. As for the second factor,
$$\begin{align}
\frac{x\cos(x/2)-2\sin(x/2)}{x^2} &= \frac{\left[x(\cos(x/2)-1)\right]+\left[\sin(x)-2\sin(x/2)\right]+\left[x-\sin(x)\right]}{x^2} \\
&= \frac{\cos(x/2)-1}{x}+\frac{2\sin(x/2)(\cos(x/2)-1)}{x^2}+\frac{x-\sin(x)}{x^2} \\
&= \frac 12\frac{\cos(x/2)-1}{x/2}+\frac 12\frac{\sin(x/2)}{x/2}\frac{\cos(x/2)-1}{x/2}+\frac{x-\sin(x)}{x^3}\cdot x
\end{align}$$
You can finish with the aforementioned limits. The limit of the second factor is $0$. So the desired limit is $1\cdot 0=0$.
A: I'll make use of some limits famously provable without the banned techniques:$$\begin{align}\frac{x\sin x+2(\cos x-1)}{x^3}&=\frac{2x\sin\frac{x}{2}\cos\frac{x}{2}-4\sin^2\frac{x}{2}}{x^3}\\&=\underbrace{\frac{4\sin^2\frac{x}{2}}{x^2}}_{\to1}\underbrace{\frac{\frac{x}{2}\cot\frac{x}{2}-1}{x}}_{\to0}.\end{align}$$The proof that $\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\tan x}{x}=1$ is famous, whence$$\lim_{x\to0}\frac{x\cot x-1}{x}=\lim_{x\to0}\frac{x-\tan x}{x\tan x}=\lim_{x\to0}\frac{x-\tan x}{x^2}=0$$because $\lim_{x\to0}\frac{x-\tan x}{x^3}=-\frac13$ is famous too (see e.g. @bjorn93's link above).
A: Whether series is allowed or not, here's the answer.
$x\sin x = x(x-x^3/3!+x^5/5!...)$ so the lowest order term is $x^2$. Simiilarly,
$2(1-\cos x)=2[1-(1-x^2/2!+x^4/4!...)]=2[x^2/2-x^4/4!...]$ so the lowest order term is $x^2$.
When you subtract the second from the first, the $x^2$ terms will cancel. The next lowest order in the numerator is $O(x^4)$. Then it is clear that the limit is $0$.
A: Note,
$$\frac{x\sin x + 2(\cos x -1)}{x^3}
=\frac{\sin x}{x}\cdot \frac {x-2\tan\frac x2}{x^2}$$
and,
$$x-2\tan\frac x2 = O(x^3)$$
Thus, 
$$\lim_{x \to 0}{\frac{x\sin{x}+2(\cos{x}-1)}{x^3}}
= \lim_{x \to 0}\frac{\sin x}{x}\cdot \frac {O(x^3)}{x^2}
=1\cdot 0=0$$
A: Let $ x\in\mathbb{R}^{*} $, observe that : $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{x-\sin{x}}{x^{2}}=\frac{x}{2}\int_{0}^{1}{\left(1-t\right)^{2}\cos{\left(tx\right)}\,\mathrm{d}t}\end{array}$} $$
Using the fact that $ \left(\forall t\in\left[0,1\right]\right),\ \left|\cos{\left(tx\right)}\right|\leq 1 $, we have : \begin{aligned} \left|\frac{x-\sin{x}}{x^{2}}\right|=\frac{\left|x\right|}{2}\left|\int_{0}^{1}{\left(1-t\right)^{2}\cos{\left(tx\right)}\,\mathrm{d}t}\right|&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-t\right)^{2}\left|\cos{\left(tx\right)}\right|\mathrm{d}t}\\&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-t\right)^{2}\,\mathrm{d}t} \end{aligned}
Which means $ \left(\forall x\in\mathbb{R}^{*}\right),\ \left|\frac{x-\sin{x}}{x^{3}}\right|\leq\frac{\left|x\right|}{6} $, and thus $ \lim\limits_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 \cdot $
Let $ x\in\mathbb{R}^{*} $, observe that : $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}=\int_{0}^{1}{\left(1-t\right)^{2}\sin{\left(tx\right)}\,\mathrm{d}t}\end{array}$} $$
Using the fact that $ \left(\forall t\in\left[0,1\right]\right),\ \left|\sin{\left(tx\right)}\right|\leq t\left|x\right| $, we have : \begin{aligned} \left|\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}\right|=\left|\int_{0}^{1}{\left(1-t\right)^{2}\sin{\left(tx\right)}\,\mathrm{d}t}\right|&\leq\int_{0}^{1}{\left(1-t\right)^{2}\left|\sin{\left(tx\right)}\right|\mathrm{d}t}\\&\leq\left|x\right|\int_{0}^{1}{t\left(1-t\right)^{2}\,\mathrm{d}t} \end{aligned}
Which means $ \left(\forall x\in\mathbb{R}^{*}\right),\ \left|\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}\right|\leq\frac{\left|x\right|}{12} $, and thus $ \lim\limits_{x\to 0}{\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}}=0 \cdot $
Hence, $$ \lim_{x\to 0}{\frac{x\sin{x}+2\left(\cos{x}-1\right)}{x^{3}}}=\lim_{x\to 0}{\left(\frac{x^{2}+2\left(\cos{x}-1\right)}{x^{3}}-\frac{x-\sin{x}}{x^{2}}\right)}=0 $$
A: As J.G. already shown it holds $$\begin{align}\frac{x\sin x+2(\cos x-1)}{x^3}&=\frac{2x\sin\frac{x}{2}\cos\frac{x}{2}-4\sin^2\frac{x}{2}}{x^3}\\&=\frac{4\sin^2\frac{x}{2}}{x^2}\frac{\frac{x}{2}\cot\frac{x}{2}-1}{x} \\ &= \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \frac{f(x) - f(0)}{x - 0} \end{align}$$
for $f(x) = x\cot(x)$
So we get by the mean value theorem: $$\lim_{x\to 0} \frac{x\sin x+2(\cos x-1)}{x^3} = 0\cdot\lim_{x\to 0} f'(x) = 0\cdot1 = 0$$
