Evaluating limit without L'Hospital rule I have the following limit:
$$
\lim_{x\rightarrow 0}{\frac{3^x+5^x-2}{2x}}
$$
The answer should be $\frac{\ln(3)+\ln(5)}{2}$. Is there a way of solving the limit without the use of L'Hospital's rule?
 A: Yes, there is.  If we are equipped with the limit
$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$
then we find that
$$\begin{align}
\lim_{x\to 0}\frac{3^x+5^x-2}{2x}&=\frac12\lim_{x\to 0}\left(\frac{3^x-1}{x}+\frac{5^x-1}{x}\right)\\\\
&=\frac{\log(3)}2\lim_{x\to 0}\frac{e^{\log(3)x}-1}{\log(3)x}+\frac{\log(5)}2\lim_{x\to 0}\frac{e^{\log(5)x}-1}{\log(5)x}\\\\
&=\frac{\log(3)+\log(5)}{2}
\end{align}$$
as was to be shown!
A: Or if we know the derivatives of the exponential functions, $(3^x)'=\ln3.3^x$ etc.,
$$\begin{align}
\lim_{x\to 0}\frac{3^x+5^x-2}{2x}&=\frac12\lim_{x\to 0}\left(\frac{3^x-3^0}{x-0}+\frac{5^x-5^0}{x-0}\right)\\\\
&=\frac12\large\left((3^x)'(0)+(5^x)'(0)\large\right)\\
&=\frac12(\ln3+\ln5)
\end{align}$$
A: Alternative approach:
Using Taylor Series, as $x \to 0$:
$3^x = e^{x\log(3)} = 1 + x\log(3) + o(x).$
$5^x = e^{x\log(5)} = 1 + x\log(5) + o(x).$
Therefore
$$\frac{3^x + 5^x - 2}{2x} = \frac{1 + x \log(3) + 1 + x\log(5) - 2 + o(x)}{2x} $$
$$= \frac{ ~\left\{ ~x \times [\log(3) + \log(5)] ~\right\} + o(x)}{x \times 2}.$$
Therefore,
$$\lim_{x \to 0} \frac{3^x + 5^x - 2}{2x} = \frac{\log(3) + \log(5)}{2}.$$
