Zadanie 8, Paper 1, maj 2024, TZ2, Matematyka AA IB HL

The initial form is 0/0. Differentiating gives $\lim_{x\to0}\frac{3\sec(3x)\tan(3x)+5\sin(5x)}{4x^3+4x}$, again 0/0. A second application yields $\lim_{x\to0}\frac{9\sec(3x)\tan^2(3x)+9\sec^3(3x)+25\cos(5x)}{12x^2+4}=34/4=17/2$.

A1: correct first numerator derivative. A1: correct first denominator derivative. M1: justified second application at a remaining 0/0 form. A1: correct second numerator derivative. A1: correct second denominator derivative. A1: 17/2, with limit notation used at least once. Further differentiation after the determinate second quotient is invalid. The second-application M1 is not awarded if a third l’Hôpital application is attempted, matching the source cap.