The Taylor expansion of ln(1-x) is -x -O(x^2). Of sin(x), x + O(x^2). This means that the top decreases as -2x.

The Taylor expansion of cos(x) is 1 - x^2/2 + O(x^4). Square that to get 1 - x^2 + O(x^4). The bottom therefore decreases as x^2. lim[x->0+](-2x/x^2) = -infinity. If you require the limit to exist from both sides, it doesn't: it's infinity as you approach 0 from below. If you require reals, not [ − ∞ , + ∞ ] , it also doesn't.

This is something that one can be expected to get in a few seconds if one is sharp and has common Taylor expansions in one's head: ln(1+x) looks like x for very small x, as does sin(x); while cos(x) looks like 1-x^2. that's all you need to know to answer the question. You don't even need to know the next term.

I do wonder, however, whether they meant to have it be ln(1+x), at which point it would exist and be trickier: the leading term in the top is the -x^2/2 from the log, while the bottom is x^2 from the cosine squared, giving an answer of -1/2. If they'd done this but messed up, it's far more plausible that Caroline would have missed the cos^2, and believing it to be simply 1-cos(x) she would have gotten x^2/2 on the bottom. This yields -1 in a much more plausible way than mistaking a minus sign for a product, which pretty much nobody ever does.