How to evaluate $\frac{2^{f(\tan x)}-2^{f(\sin x)}}{x^{2}f(\sin x)}$ as $x \to 0$? If $f(x+y)=f(x)+f(y)$ for all real values of $x,y$.
Given $f(1)=1$
How to evaluate $$\lim_{x \to 0} \frac{2^{f(\tan x)}-2^{f(\sin x)}}{x^{2}f(\sin x)}$$
 A: Hint: If $f(x + y) = f(x) + f(y)$, then $$f(0 + 1) = f(0) + f(1).$$ Therefore, $f(0) = f(0 + 1) - f(1)$ and thus $f(0) = 1 - 1 = 0$. So $f(0) = 0$.
A: We have $f(2x)=f(x)+f(x)=2f(x)$ and $f(3x)=f(2x)+f(x)=3f(x)$ 
Also we have $f(1)=f(0+1)=f(0)+f(1)\Rightarrow f(0)=0$  
Also $f(0)=0=f(x+(-x))=f(x)+f(-x)\Rightarrow f(-x)=-f(x)$  
Then $f(-2x)=f(-x)+f(-x)=-f(x)-f(x)=-2f(x)$
and $f(-3x)=f(-2x-x)=f(-2x)+f(-x)=-2f(x)-f(x)=-3f(x)$
So by induction we will have:
$$\forall k\in\mathbb Z:f(kx)=kf(x)$$
so we will have $f(\frac{m}{m})=mf(\frac{1}{m})\Rightarrow f(\frac{1}{m})=\frac{1}{m}f(1)\;\forall m\in\mathbb Z$
Then for rational numbers we will have
$$f(r)=f(\frac{m}{n})=mf(\frac{1}{n})=m\times\frac{1}{n}f(1)=\frac{m}{n}f(1)=rf(1)\; \forall r\in\mathbb Q$$
We now that for any irrational $x$ we have a sequence of rational numbers $r_n$ such that $\lim_{n\to \infty}r_n=x$ so from the continuity of $f$ we will have:
$$f(x)=f(\lim_{n\to\infty}r_n)=\lim_{n\to\infty}f(r_n)=\lim_{n\to\infty}r_nf(1)=f(1)\lim_{n\to\infty}r_n=xf(1)$$
So because the proof is true for rational and irrational numbers we will have:
$$f(x)=xf(1)=x\times 1=x\quad\forall x\in\mathbb R$$
So if $f$ is continuous then we have $f(x)=x$ for all $x$ so:
$$\lim_{x \to 0} \frac{2^{f(\tan x)}-2^{f(\sin x)}}{x^{2}f(\sin x)}=\lim_{x\to 0}\frac{2^{\tan x}-2^{\sin x}}{x^2 \sin x}=\lim_{x\to 0}\frac{e^{\tan x\, \ln2}-e^{\sin x\, \ln2}}{x^2\sin x}$$  
From the Maclaurin series we have $x\to 0:\sin x\simeq x$
Also from the Maclaurin series we have $x\rightarrow0\Rightarrow e^x\simeq1+x$
Substituting $x$ by $\sin x\,\ln2$ and $\tan x\,\ln2$ because both of them tend t0 $0$ when $x\rightarrow 0$ yields:
$$=\lim_{x\to 0}\ln2\frac{\tan x-\sin x}{x^3}$$
Then if we write Maclaurin series up to the third derivative for $\sin x$ and $\tan x$ we have:
$x\to 0:\sin x\simeq x-\frac{x^3}{6}\quad,\quad \tan x\simeq x+\frac{x^3}{3}$
so we will have
$$=\lim_{x\to 0}\ln2\frac{x+\frac{x^3}{3}-(x-\frac{x^3}{6})}{x^3}=\lim_{x\to 0}\frac{\ln2}{2}\frac{x^3}{x^3}=\frac{\ln2}{2}$$
