Evaluate: $\lim_{x\to5}\left(\dfrac{\log{(x+5)}-\log{(5-x)}}{x-5}\right)$
If we put $x=5$, we get $\dfrac{\infty}{0}$ form, which is not an indeterminate form, right?
So, I was tempted to write 'not defined' as answer. Yet I half-heartedly applied L'Hospital but didn't get any answer. I expanded $\log$ series. Still no answer.
Even after looking at the following solution, I am not able to understand it.
$$\lim_{x\to5}\left(\frac{\log(x+5)-\log(5-x)}{x-5}\right)\\=\lim_{x\to5}\left(\frac{(\log(x+5)-\log5)-(\log(5-x)-\log5)}{x-5}\right)\\=\lim_{x\to5}\left(\left(\frac{\log(x+5)-\log5}{x-5}\right)-\left(\frac{\log(5-x)-\log5}{x-5}\right)\right)\\=\lim_{x\to0}\left(\left(\frac{\log(x+5)-\log5}{x-5}\right)+\left(\frac{\log(5-x)-\log5}{5-x}\right)\right)\\=\frac15+\frac15=\frac25$$
I don't understand why and how the limit changed to $x\to0$.
Best Answer
Write $x:=5-\mathrm e^{-t}\;\,(t>0)$. Then $$\lim_{x\to5}\frac{\log(x+5)-\log(5-x)}{x-5}=\lim_{t\to\infty}\frac{\log(10-\mathrm e^{-t})+t}{\mathrm e^{-t}}=\infty.$$ So there is no limit. The change of the $x$ limit to $x\to0$ in the solution you cite is just plain wrong.