Infinity as a limit in an indeterminate form

Question

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?

Answer 1

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.

Answer 2

You don't "turn" an expression into an indeterminate form. It is an indeterminate form until you have computed its limit.

You did recognize an indeterminate form of the type $\dfrac00$, now your task is to evaluate it. L'Hospital is a way.