Application of l'Hopital's Rule in a past paper In an old exam paper the following solution is given:
$$\lim_{r \to 0}\cfrac{\sin\left(\cfrac{\mu_1\cdot r}{R}\right)}{\mu_1\cfrac{r}{R}}\overset{l'Hopital}{=}\cfrac{\cfrac{1}{R}\cos\left(\cfrac{\mu_1\cdot r}{R}\right)}{\mu_1\cfrac{1}{R}}=\frac{1}{\mu_1}$$
My understanding of l'Hopital is that the result should be $1$:
$$\lim_{r \to 0}\cfrac{\sin\left(\cfrac{\mu_1\cdot r}{R}\right)}{\mu_1\cfrac{r}{R}}=\cfrac{\cfrac{\mu_1}{R}\cos\left(\cfrac{\mu_1\cdot r}{R}\right)}{\mu_1\cfrac{1}{R}}=1$$
Is this a mistake or have I misunderstood something?

You are right. One might also substitute $\frac{\mu_1r}{R}=: h$ and then ends up with $\lim_{h\to 0}\frac{\sin h}h=1$
