Let $a>0$ and $$g(x) = \begin{cases} \dfrac {1 - a^x + xa^x \ln a} {a^x x^2}, & x<0 \\ \dfrac {2^x a^x - x \ln 2 - x \ln a - 1} {x^2}, & x>0 \end{cases} .$$ Without using L'Hopital's rule or power series, find the value of $a$ and $g(0)$ so that the function $g(x)$ is continuous at $x = 0$.
I thought about it, but do not know where to begin.
Hint: We know $(a^x)' = (\ln a) a^x.$ Thus the derivative of $a^x$ at $x=0$ is $\ln a.$ From the definition of the derivative we then have $a^x= 1 +(\ln a) x +o(x)$ as $x\to 0.$ (No L'Hopital or power series results were used.)
It follows from the FTC that
$$a^x = 1 +\int_0^x (\ln a) a^t \, dt = 1 +\int_0^x \ln a (1 +(\ln a) t +o(t))\,dt = 1 + (\ln a) x + (\ln a)^2x^2/2 + o(x^2).$$
Again, no Taylor or L'Hopital was used–just the definition of the derivative and the FTC.
This should allow you to solve the problem.