Need to find limit, given function Given that


*

*$f(x+y) = f(x) + f(y)$ for all real $x, y$

*$f(1) = 1$. 


To show that $$\lim_{x \to 0}\frac{f(x)}{x}=1$$ 
My intuition says that (1) gives us $f(x) = cx$ for some real $c$ ; but I am unable to argue why that should be.
Edit:
My actual question is to evaluate below :
$$\lim_{x \to 0}\frac{2^{f(\tan x)} - 2^{f(\sin x)} }{x^2 f(\sin x) } $$
Which evaluates assuming $\frac{f(x)}{x}$ approaches $~1~$ as $~x~$ approaches to $~0~$.
 A: There exist discontinous functions satisfying the given conditions. Such functions are not bounded in any neighborhood of $0$ so they do not satisfy $\lim _{x \to 0} \frac {f(x)} x=1$. So the result, as stated, is false. However, if $f$ is continuous then the first condition implies that $f(x)=cx$ for some $c$ and the second condition implies that $c=1$ so we are done. A proof of the fact that if $f$ is continuous then  $f(x)=cx$ for some $c$ can be found easily by searching for Cauchy's Functional Equation. 
A: First notice that from $(1)$ you can conclude that $f(\frac{1}{n}) = \frac{f(1)}{n}$ which by $(2)$ equals $\frac{1}{n}$. From this you can conclude that if the limit in the question exists, it must be $1$.
Unfortunately, the limit may not exists without further assumptions on $f$.
Consider the set $\{1,\pi,\pi^2,...\}$. This is a set of independent real numbers over $\mathbb{Q}$ and so we can extend it to a basis of $\mathbb{R}$ over $\mathbb{Q}$.
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be the unique linear map which takes $\pi^k$ to $4^k$ and every other element in the basis to zero.
Then $f(1) = f(\pi^0) = 4^0 = 1$ so $(1),(2)$ are satisfied.
However $f(\pi^k) = 4^k$ and so $f( \frac{1}{4^k} \pi^k)/\frac{1}{4^k}\pi^k = (\frac{4}{\pi})^k$ does not converge to $1$ as $k$ goes to infinity.
Edit: As mentioned by @kavi Rama Murty, the conclusion becomes true if you assume that $f$ is continuout. In fact, given assumption $(1)$ it is enough to assume that $f$ is Lebesgue measurable (as any measurable homomorphism is continuous). In particular this means that you can't construct a counter-example without using the axiom of choice.
