If f(x + y) = f(x) + f(y) for all x, y in R a | vedclass

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(A) Given the Cauchy functional equation $f(x + y) = f(x) + f(y)$,the solution is of the form $f(x) = cx$.Since $f(1) = 1$,we have $c(1) = 1$,so $c =

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$\log_e 2$

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(A) Given the Cauchy functional equation $f(x + y) = f(x) + f(y)$,the solution is of the form $f(x) = cx$.Since $f(1) = 1$,we have $c(1) = 1$,so $c = 1$. Thus,$f(x) = x$.The expression becomes $\lim_{x 	o 0} \frac{2^{	an x} - 2^{\sin x}}{	an x - \sin x}$.Let $u = 	an x$ and $v = \sin x$. As $x 	o 0$,$u 	o 0$ and $v 	o 0$.The expression is $\lim_{x 	o 0} \frac{2^{	an x} - 2^{\sin x}}{	an x - \sin x} = \lim_{x 	o 0} \frac{2^{\sin x}(2^{	an x - \sin x} - 1)}{	an x - \sin x}$.Using the standard limit $\lim_{t 	o 0} \frac{a^t - 1}{t} = \ln a$,where $t = 	an x - \sin x$,we get:$= \lim_{x 	o 0} 2^{\sin x} \cdot \lim_{x 	o 0} \frac{2^{	an x - \sin x} - 1}{	an x - \sin x} = 2^0 \cdot \ln 2 = \ln 2 = \log_e 2$.

If $f(x + y) = f(x) + f(y)$ for all $x, y \in R$ and $f(1) = 1$,then find the value of $\lim_{x \to 0} \frac{2^{f(\tan x)} - 2^{f(\sin x)}}{f(\tan x) - f(\sin x)}$.

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