Prove that $\lim_\limits{x\to 0}{f(x)}=0$ 
Let $f:\mathbb{R}\mapsto \mathbb{R}$ be a function such that: $$2\cdot f(x)-\sin(f(x))=x, \forall x\in \mathbb{R}$$ Prove that
  $\lim_\limits{x\to 0}{f(x)}=0$.

I think I need to use the sandwich theorem, so I have found its first part, but not the second as follows:
$\forall x\in \mathbb{R}$ and $\forall f(x)\in \mathbb{R}$, it is: $$\sin(f(x)) \leq f(x) \Rightarrow 2\cdot f(x)-x \leq f(x) \Rightarrow f(x) \leq x$$
So, now I have to figure out how to have a perfect sandwich theorem application. Any hint? (I haven't yet been taught derivatives and differentiability)
 A: Hint:
Write the identity as:
$$f(x)=\frac{x+\sin f(x)}2$$
Now, apply the triangle inequality and that $|\sin t|\le |t|$ for any $t\in\Bbb R$:
$$0\le|f(x)|\le\frac{|x|+|f(x)|}2$$
A: Suppose the contrary. Then there exists $\varepsilon>0$ such that, for all positive integers $n$, there is $x_n$ with $0<|x_n|<\frac{1}{n}$ and $|f(x_n)|>\varepsilon$.
The given identity yields
$$
2f(x_n)-\sin f(x_n)=x_n
$$
Now it's not difficult to show that $|2t-\sin t|\ge|t|$ (from the standard inequality $\sin t\le t$ when $t\ge0$, equality holding only for $t=0$). So we have
$$
|x_n|=|2f(x_n)-\sin f(x_n)|\ge |f(x_n)|>\varepsilon
$$
which is a contradiction as soon as $1/n<\varepsilon$.
A: Hint: not quite what you had in mind, but it certainly suffices to show that the function $g(x)=2x-\sin(x)$ has a continuous inverse.
In order to show this, it suffices to note that $g$ is differentiable with $g'(x)>0$
A: If you assume the contrary, i.e $f(x)\rightarrow a\neq 0$, then for small enough $|x|<\delta$ you have that $|f(x)-a|<\epsilon$, so $f(x)\neq 0$ for such $x$ (choose $\epsilon$ small enough, so that $0\notin (a-\epsilon,a+\epsilon)$) and you can divide by $f(x)$ both sides of the equation. Letting $x\rightarrow 0$ you get $2-\frac{sin(a)}{a}=0\Rightarrow 2a=sin\,a$ which has only the zero solution.
Note: $f(x)$ can not tend to $\infty$. 
