$ f \colon \mathbb R \to \mathbb R$ be a function such that $f(x+y)= f(x) f(y)$ I came across the following problem that says:  

Let $f \colon \mathbb R \to \mathbb R$ be a function such that $ f(x+y)= f(x) f(y) ,\forall x,y \in \mathbb R$ and $f(x)=1+x g(x); $ where $\lim\limits_{x \to 0} g(x)=1.$ Then the function $ f(x) $ is which of the following?  
  
  
*
  
*$e^x$  
  
*$2^x$  
  
*A non-constant polynomial  
  
*equal to $1$ $\forall x \in \mathbb R$.   
  

EDIT: My Attempt:  We see that $\lim\limits_{x \to 0} \dfrac{f(x)-f(0)}{x-0}=1 $ (By Calculating and putting the value of $f(x)$) and that means $f'(0)=1.$ Also putting $y=0$ in the relation $ f(x+y)= f(x) f(y) ,\forall x,y \in \mathbb R$,I get $f(0)=1.$ So,I can eliminate options $2$ and $4.$ Am I going in the right direction?
Can someone point me in the right direction? Thanks in advance for your time.
 A: Note that $f(0) = 1$.
Step 1. $f$ is differentiable everywhere and satisfies $f'(x) = f(x)$. This is because
$$ f'(x) = \lim_{h\to0}\frac{f(x+h) - f(x)}{h} = f(x)\lim_{h\to0}\frac{f(h) - 1}{h} = f(x)\lim_{h\to0}g(h) = f(x). $$
Step 2. $f(x) = e^x$. This follows from the observation that
$$ \left(f(x)e^{-x}\right)' = f'(x)e^{-x} - f(x)e^{-x} = 0. $$
A: Hint 1: What's another way of writing the derivative of $f$ at an arbitrary point $x$? Once you rewrite it, you should use the functional equations given to you to simplify.
Hint 2: What other function(s) do you know that satisfies the differential equation you created?
Hint 3: Can you find $f(0)$?
A: You correctly eliminated options $2$ and $4$. To eliminate $3$, compare degrees in $f(2x)=f(x)^2$.
A: Well, its $e^x.$ 
$f(x+y)=f(x)f(y)$ $\implies$ This is a power function. Therefore answer has to be either $2^x$ or $e^x.$
Let $f(x)= 2^x$
$2^x=1+xg(x)$
$g(x)= \frac{2^x-1}{x}$
We have
$$\lim_{x \to 0} \frac{2^x-1}{x}= \log2$$ which is not equal to one.
Whereas,
$$\lim_{x \to 2} \frac{e^x-1}{x}= 1$$
