There is this limit
$$\lim_{x\to1}\frac{\ln x}{\left(x-1\right)^{2}}$$
for which I built a graph and I know the answer, so the question is not about how to compute it, but about my observation that when $x$ approaches $1$ from the right, the $\ln x = 0^+ $, but when $x$ approaches $1$ from the left, the $\ln x = 0^-$.
Based on that fact, even though it's an an indeterminate form $0 : 0$, can I already say that the limit does not exist or is that not enough and I need to turn it into a determinate form? And if I have to turn it into a determinate form, how can I do that?
Best Answer
To evaluate indetermination just plug in the limit value for numerator and denominator to see that it is in the form $\frac 0 0$.
For the solution we can avoid l'Hospital by $x-1=t \to 0$ to obtain
$$\frac{\ln x}{\left(x-1\right)^{2}}=\frac{\ln (1+t)}{t^{2}}=\frac1t\frac{\ln (1+t)}{t}$$
with $\frac{\ln (1+t)}{t}\to 1$ therefore we can conclude that limit doesn't exist.