$\lim_{x\rightarrow0} \frac{a\log(1+x)-\log(1+ax)}{x^2}$ without L'Hôpital's rule

Question

Without using L'Hôpital's, what is the following limit?

$$\lim_{x\rightarrow0} \frac{a\log(1+x)-\log(1+ax)}{x^2}$$

Answer 1

Since $\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + o(x^4)$, as $x\to 0$, we have

$\log(1+ax) = ax - \frac{a^2x^2}{2}+ \frac{a^3x^3}{3} + o(x^4)$.

So,

$\frac{a\log(1+x)}{x^2} = a(\frac{1}{x} - \frac{1}{2} + o (x))$

and

$\frac{\log(1+ax) }{x^2}= \frac{a}{x} - \frac{a^2}{2} +o(x).$

Then,

$\lim\limits_{x\to0}\frac{a\log(1+x) - \log(1+ax)}{x^2} = \lim\limits_{x\to0}\left[(\frac{a}{x}-\frac{a}{2}+o(x)) - (\frac{a}{x} - \frac{a^2}{2} + o(x) ) \right] = \frac{a^2-a}{2}. $

Answer 2

Since $\ln$ is continous function on its domain the limit asked, if exists, is $\ln L$ where $$L=\lim_{x\rightarrow 0} \left(\frac{(1+x)^a}{1+ax}\right)^{\frac{1}{x^2}}.$$ Now, we compute $$\frac{(1+x)^a}{1+ax}=(1+ax+\frac{a^2-a}{2}x^2+O(x^3))(1-ax+a^2x^2+O(x^3))=1+\frac{a^2-a}{a}x^2+O(x^3).$$ So, $$L=\lim_{x\rightarrow 0}\left(1+\frac{a^2-a}{2}x^2+O(x^3)\right)^{\frac{1}{x^2}}.$$ Since, $\lim_{x\rightarrow 0}(1+O(x^3))^{\frac{1}{x^2}}=1$, by comparision, we can conclude that $$L=\lim_{x\rightarrow 0}\left(1+\frac{a^2-a}{2}x^2\right)^{\frac{1}{x^2}}.$$ Now, let $x^2=\frac{2u}{a^2-a}$ then $$L=\left(\lim_{x\rightarrow 0}(1+u)^{\frac{1}{u}}\right)^{\frac{a^2-a}{2}}=e^\frac{a^2-a}{2}.$$ For the last equality, see https://www2.math.uconn.edu/~hurley/math116/section4_docs/Handouts/elimit.pdf.

Hence the answer is $\ln L=\frac{a^2-a}{2}.$