Evaluate $\lim\limits_{x\to 0 } \frac{ e^{\frac{\ln(1+ax)}{x}} - e^{\frac{a\ln(1+x)}{x}}}{x}$. Problem
Evaluate
$$\lim_{x\to 0 } \frac{ e^{\frac{\ln(1+ax)}{x}} - e^{\frac{a\ln(1+x)}{x}}}{x}.$$
Solution
For  convenience, denote $u(x)=\dfrac{\ln(1+ax)}{x}$ and $v(x)=\dfrac{a\ln(1+x)}{x}.$ Notice that $u(x),v(x) \to a$ as $x \to 0.$Hence,
\begin{align*}
\lim_{x\to 0 } \frac{ e^{\frac{\ln(1+ax)}{x}} - e^{\frac{a\ln(1+x)}{x}}}{x}&=\lim_{x\to 0 } \frac{e^{v(x)}(e^{u(x)-v(x)}-1)}{x}\\&=\lim_{x \to 0}\left(e^{v(x)}\cdot\frac{e^{u(x)-v(x)}-1}{u(x)-v(x)}\cdot \frac{u(x)-v(x)}{x}\right)\\&=\lim_{x \to 0}e^{v(x)}\cdot\lim_{x \to 0}\frac{e^{u(x)-v(x)}-1}{u(x)-v(x)}\cdot \lim_{x \to 0}\frac{u(x)-v(x)}{x}\\&=e^a \cdot 1 \cdot\lim_{x \to 0}\frac{\ln(1+ax)-a\ln(1+x)}{x^2}\\&=e^a \cdot\lim_{x \to 0}\dfrac{ax-\dfrac{1}{2}a^2x^2+\mathcal{O}(x^2)-a\left(x-\dfrac{1}{2}x^2+\mathcal{O}(x^2)\right)}{x^2}\\&=e^a \cdot\lim_{x \to 0}\dfrac{-\dfrac{1}{2}a^2x^2+\dfrac{1}{2}ax^2+\mathcal{O}(x^2)}{x^2}\\&=\frac{1}{2}e^a(a-a^2). \end{align*}
Please correct me if I'm wrong! Hope to see other solutions.
 A: The OP used series in the work to arrive at the answer. Another approach is to use the definition of the derivative. First it is easy to observe that $\lim\limits_{x\to 0 }\ln\frac{(1+ax)}{x}=a$ and  $\lim\limits_{x\to 0 }a\ln\frac{(1+x)}{x}=a$, verified through L'Hospital's Rule. 
So $\lim\limits_{x\to 0 } \frac{ e^{\frac{\ln(1+ax)}{x}} - e^{\frac{a\ln(1+x)}{x}}}{x}$ can be written as  $\lim\limits_{x\to 0 } \frac{ e^{\frac{\ln(1+ax)}{x}}-a}{x-0}$ - $\lim\limits_{x\to 0 } \frac{ e^{\frac{a\ln(1+x)}{x}}-a}{x-0}$
Here we see twice the definition of the derivative of two functions (e-powered to the ln...). So essentially this means that we have to take the derivative of those e-power functions and "plug in" $x=0$. In both cases that isn't too easy in the sense that we will get an indeterminate form (0/0) but the limits in my introduction might be helpful with that. The derivative of $ e^{\frac{\ln(1+ax)}{x}}$ is $ e^{\frac{\ln(1+ax)}{x}}\times[\frac{ln(1+ax)}{x}]'$ (Chain Rule). The derivative of that second part (after simplifying) is $\frac{ax-(1+ax)ln(1+ax)}{(1+ax)x^2}$. Putting in $x=0$ gives an indeterminate form but with Hospital's Rule (or some other manipulation, like with Standard limits), this yields $-\frac{1}{2}a^2$. With the e-power upfront: $-\frac{1}{2}e^aa^2$ which is one part of your answer. The other derivative goes in a very similar fashion and will give you the other part of your answer. Hope this helps.
