L'Hopital's rule step by step: limit of arcsin(3x)/tan(5x) as x→0

The question

The solution

1. 01Substitute $x = 0$ into the top and the bottom before doing anything else: $\frac{\arcsin 0}{\tan 0} = \frac{0}{0}$. This is not an answer, it is the permission to use the rule, and writing it down earns a mark on its own.
2. 02State the rule: differentiate the top and the bottom separately, then take the limit again, $\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}$.
3. 03Do not use the quotient rule $\frac{f'g - fg'}{g^{2}}$. L'Hôpital's rule does not differentiate the fraction as one object, it differentiates the top on its own and the bottom on its own. This is the line that costs the mark.
4. 04Differentiate the top: $\frac{d}{dx}\arcsin 3x = \frac{3}{\sqrt{1 - 9x^{2}}}$. The 3 on top comes from the chain rule and is the most forgotten factor in the question.
5. 05Differentiate the bottom: $\frac{d}{dx}\tan 5x = 5\sec^{2} 5x$. The 5 inside comes out and multiplies. An answer of 1 instead of $\frac{3}{5}$ almost always means the inner 3 and 5 went missing.
6. 06Put the two derivatives back over each other and let $x \to 0$: $\lim_{x\to 0}\frac{3/\sqrt{1-9x^{2}}}{5\sec^{2}5x} = \frac{3/1}{5\cdot 1}$, because the square root becomes $\sqrt{1}$ and $\sec 0 = 1$.
7. 07Read off the result: $\frac{3}{5} = 0.6$. This matches the height the graph was heading towards near zero, and if the algebra and the graph disagree, the algebra is usually wrong.
8. 08Check with small angles: $\arcsin 3x \approx 3x$ and $\tan 5x \approx 5x$, so near zero the fraction is $\frac{3x}{5x} = \frac{3}{5}$.
9. 09If substituting after differentiating gives $\frac{0}{0}$ again, the rule can be applied again. It must not be applied when the substitution already gives a number.

Answer
The limit is 3/5 = 0.6, found by l'Hôpital's rule after checking that the form is 0/0, then differentiating top and bottom separately.