Let be $f:\mathbb R \to \mathbb R$ be a function satisfying $f(x+y)=f(x).f(y)$. Which one of the following are correct? 
Let be $f:\mathbb R \to \mathbb R$ be a function satisfying
  $f(x+y)=f(x)\cdot f(y)$ and $\displaystyle\lim_{x \to 0}f(x) = 1$. which of the following
  are necessarily true?
  
  
*
  
*$f$ is strictly increasing
  
*$f$ is either constant or bounded
  
*$f(rx)=f(x)^r, \forall r \in \mathbb Q$
  
*$f(x)\ge 0, \forall x \in \mathbb R$

$f(2x)=f(x)^2$
We can prove by induction that
$\forall m\in \mathbb Z f(mx)=f(x)^m$
$f(\frac{nx}{n})=f(\frac{x}{n})^n \implies f(\frac{mx}{n})= f(x)^\frac{m}{n}$. hence 3. is correct.
$e^x$ satisfying above condition, So, 1,4 are right. Hence, 1,3,4 are correct answers. But answer given in the answer key were 3 and 4. I am not able to give any counter example for 1. 
 A: If $f$ vanishes somewhere say at $a$ then $f(b) =f(a) f(b-a) =0$ and hence $f$ is identically zero which contradicts $\lim_{x\to 0}f(x)=1$. Thus $f(x) \neq 0$ for all $x\in\mathbb{R} $ and then $f(x) =f(x/2)f(x/2)>0$ for all $x\in\mathbb{R} $. This proves point 4.
Next we can see that $$\lim_{h\to 0}f(x+h)=\lim_{h\to 0}f(x)f(h)=f(x)\cdot 1=f(x)$$ so that $f$ is continuous everywhere. Now it is easy to show using algebra that $f(rx) =\{f(x) \} ^r$ for all $x\in\mathbb {R}, r\in\mathbb {Q} $ (first prove this for the case when $r$ is a positive integer, then extend it to the case when $r$ is a negative integer and finally extend to the case when $r$ is rational). Point 3 is done. By continuity the relation holds for all $r\in\mathbb{R} $ and hence $f(x) =\{f(1)\} ^x=a^x$ where $a=f(1)$.
Thus the given function is of the form $f(x) =a^x$ where $a>0$. If $a=1$ then the function is constant, if $a>1$ then function is strictly increasing and if $0<a<1$ then it is strictly decreasing. This removes point 1. For point 2 note that the function is unbounded if $a\neq 1$.
A: Your function is not $e^x$ ! You have for example that $f(x)=1^x$ satisfy all condition and is constant.
A: What about $f(x)=1$ as a counter example for case 1 ? 
