Answer
(a)
$L(ab)=\lim\limits_{x \to 0}\frac{(ab)^{x}-1}{x}=\lim\limits_{x \to 0}\frac{a^{x}(b^{x}-1)+(a^{x}-1)}{x}=\lim\limits_{x \to 0}\frac{a^{x}(b^{x}-1)}{x}+\frac{a^{x}-1}{x}=\lim\limits_{x \to 0}\frac{b^{x}-1}{x} + \lim\limits_{x \to 0}\frac{a^{x}-1}{x}=L(b)+L(a)$
(b)
$L(12)=L(3\times4)=\lim\limits_{x \to 0}\frac{(4\times3)^{x}-1}{x}=\lim\limits_{x \to 0}\frac{4^{x} (3^{x}-1) }{x}+\lim\limits_{x \to 0}
\frac{4^{x}-1}{x}=L(3)+L(4)$
Work Step by Step
(a) . ab>0, thus $L(ab)=\lim\limits_{x \to 0}\frac{ab^{x}-1}{x}$ exist.
And using : $ab^{x}-1=a^{x}(b^{x}-1)+(a^{x}-1)$ , we get
$
L(ab)=\lim\limits_{x \to 0}\frac{a^{x}(b^{x}-1)+(a^{x}-1)}{x}=\lim\limits_{x \to 0}(a^{x}.\frac{b^{x}-1}{x}+\frac{a^{x}-1}{x})
$
We know that : $\lim\limits_{x \to 0}a^{x}=1$ and $\lim\limits_{x \to 0}\frac{b^{x}-1}{x}=L(b)$
Thus $\lim\limits_{x \to 0}a^{x}.\frac{b^{x}-1}{x}=L(b) ...(*)$
And we have $\lim\limits_{x \to 0}\frac{a^{x}-1}{x}=L(a) ...(**)$
Hence $L(ab)=\lim\limits_{x \to 0}(a^{x}.\frac{b^{x}-1}{x}+\frac{a^{x}-1}{x})=L(b)+L(a)$
(b) . We have
$ L(3)=\lim\limits_{x \to 0}\frac{3^{x}-1}{x}$
$L(4)=\lim\limits_{x \to 0}\frac{4^{x}-1}{x}$
In the other hand we have
$L(12)=L(4\times3)=\lim\limits_{x \to 0}\frac{(4\times3)^{x}-1}{x}=\lim\limits_{x \to 0}\frac{4^{x}(3^{x}-1)+(4^{x}-1)}{x}=\lim\limits_{x \to 0}(4^{x}.\frac{3^{x}-1}{x}+\frac{4^{x}-1}{x})=\lim\limits_{x \to 0}4^{x}.\frac{3^{x}-1}{x}+\lim\limits_{x \to 0}\frac{4^{x}-1}{x}$
Using $(*)$ from the previous question we get
$\lim\limits_{x \to 0}4^{x}.\frac{3^{x}-1}{x}=L(3)$
Hence
$L(12)=L(3)+L(4)$
