Is $\dfrac{1}{\frac{0}{0}-1}$ an indeterminate form? I thought only $\dfrac 00,\,\dfrac\infty\infty$ and any form that can be represented in those two are indeterminate. Moreover, how do we know if a form is indeterminate?
P.S.
For people who says that this question doesn't reflect OP's effort, I came through it when calculating $$\lim_{t\rightarrow 0} \frac{1}{\frac{2\sin^2\frac{t}{2}\sin t}{(1-\cos t)(t-\sin t)}-1}.$$ I went to a Wikipedia page on indeterminate forms and there wasn't any mention of it.
Let us consider the main purpose of this topic to solve the limit and discuss the method of solving it.
As I mentioned in the comments, the use of L'Hopital's rule is not determined by some or other "indeterminate forms". The theorem has very specific assumptions. Furthermore, I do not know what it means to be "completely indeterminate" and how that would specify the condition of being "indeterminate".
Suppose we needed to find $$\lim _{t\to 0} \frac{f(t)}{g(t)} $$ If $\lim _{t\to 0} f(t) = \lim_{t\to 0} g(t) =0$ OR $\lim_{t\to 0} f(t) =\pm\infty$ and $\lim _{t\to 0} g(t) =\pm\infty$, then $$\lim _{t\to 0} \frac{f(t)}{g(t)} = \lim _{t\to 0} \frac{f'(t)}{g'(t)} $$ Of course, $f$ and $g$ must be differentiable around $t=0$, but that is not an obstacle for this particular problem.