Solving Functional Equations by Limits ... Let $f(x)$ be a continuous function and satisfying the equation : 
$f(2x) - f(x) = x$. Given $f(0)=1$ ; Find $f(3)=?$
My teacher solves this as : 
$$f(x) - f(x/2) = x/2$$
$$f(x/2) - f(x/4) = x/4$$
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$$f(x/2^{n-1})  - f(x/2^{n}) = x/ 2^{n-1}$$
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Add them up : And let $n\rightarrow \infty$
$$f(x)  - f(x/2^{n}) =(x/2)/(1-(1/2))$$
Thus $f(x)  - f(0) = x$.
Thus $f(x) = x+1$. Thus $f(3)= 4$.
However my problem is that I didn't find it intuitive ; I didn't understand how to get such an idea.
So is there an alternate way to go about this problem ? 

I would say, 
search function $f(x) = x+\mu(x), \quad \mu(x)$ continuous for $x\in R,\,\mu(0)=1$:
$f(2x)-f(x)=x\Rightarrow 2x + \mu(2x) - x - \mu(x)=x\Rightarrow \mu(2x)=\mu(x)$
$\Rightarrow \mu(x) \equiv$ const $= \mu(0) = 1$
$\Rightarrow f(x)=x+1\Rightarrow f(3)=4$
