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.

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.