I tutor for AP Calc AB, it's been a couple years since I last took calculus though. My student had a question with a limit using L'Hopital's Rule, with $\infty/\infty$.
The question was: $$\lim _{x \to \frac{\pi}{2}^-} \frac{\tan{x}}{3(2x-\pi)^{-1}}$$
This yields $\infty/\infty$, so after taking the derivative of the numerator and denominator: $$\lim _{x \to \frac{\pi}{2}^-} \frac{\sec^{2}{x}}{-\frac{6}{(2x-\pi)^2}}$$
Not sure what to do from here. Symbolab gives $-\frac{2}{3}$ as the answer.
Rewrite as $\displaystyle \lim_{x \to \pi/2{}^-} f(x) \times g(x) ~: ~f(x) = \sin(x), ~ g(x) = \frac{2x - \pi}{3\cos(x)}.$
In Real Analysis, you have that if the limit of each of the individual factors exist, then the limit of the product is the product of the limits.
Clearly, $\displaystyle \lim_{x \to \pi/2{}^-} f(x) = 1.$
Then
$$\lim_{x \to \pi/2{}^-} g(x) = \lim_{x \to \pi/2{}^-} \frac{2x - \pi}{3\cos(x)}. \tag1 $$
From L'Hopital's rule, it is immediate that
$$\lim_{x \to \pi/2{}^-} g(x) = -\frac{2}{3}.$$