I can calculate this easily using L'Hopital Rule. Can anyone give me some pointers on how to do this without using L'Hopital?
$$\lim_{x \to a} \frac{\log(x-a)}{\log(e^x-e^a)}$$
I tried substitution by $n = x+a$ and then $\lim_{n \to 0}$ but that didn't work.
You need to substitute $n=x-a$ instead. Then, note that the required limit is $$\lim_{n\to 0}\frac{\log n}{\log(e^a\times e^{x-a}-e^a)}$$ Then, you can factor, and proceed with the known identity.
Hope this helps. Ask anything if not clear :)
It is always easier to work close to $0$. So, making $x=t+a$ $$\lim_{x \to a} \frac{\log(x-a)}{\log(e^x-e^a)}=\lim_{t \to 0} \frac{\log(t)}{\log(e^a(e^t-1))}=\lim_{t \to 0} \frac{\log(t)}{a+\log(e^t-1)}$$ Now, using Taylor expansion $$e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+O\left(t^4\right)$$ $$\log(e^t-1)=\log\big[t+\frac{t^2}{2}+\frac{t^3}{6}+O\left(t^4\right)\big]$$ $$\log(e^t-1)=\log (t)+\frac{t}{2}+\frac{t^2}{24}+O\left(t^3\right)$$ $$\frac{\log(t)}{\log(e^a(e^t-1))}=\frac{\log (t)}{a+\log (t)}-\frac{t \log (t)}{2 (a+\log (t))^2}+O\left(t^2\right)$$ which shows that the limit is $1$ and also how it is approached.
