Zadanie 9, Paper 1, maj 2025, TZ3, Matematyka AA IB HL

The form is 0/0. Applying l’Hôpital gives $\lim_{x\to0}[\sin3x+3x\cos3x]/[2\sin2x]$, still 0/0. A second application gives $\lim_{x\to0}[6\cos3x-9x\sin3x]/[4\cos2x]=6/4=3/2$. Alternatively use $x\sin3x=3x^2-\frac92x^4+O(x^6)$ and $1-\cos2x=2x^2-\frac23x^4+O(x^6)$, cancel x² and take the limit. Another exact route uses $1-\cos2x=2\sin^2x$ and fundamental sine limits. After one l’Hôpital step, another valid route rewrites the quotient as $\frac34\frac{\sin(3x)/(3x)+\cos(3x)}{\sin(2x)/(2x)}$, which tends to $\frac34(1+1)=3/2$.

For l’Hôpital: R1 identify 0/0 before applying the rule; A1 first differentiated quotient; M1 apply it again only while the form remains indeterminate; A1 second quotient; M1 substitute the limiting input; A1 result 3/2. Later marks are independent of the first reason mark. For series: M1 expand either function with two terms; A1 both correct expansions; M1 identify/cancel x²; A1 simplified ratio; M1 take the limit; A1 result. For a trig-identity route: M1 use a suitable identity or conjugate; A1 correct rewritten expression; A1 reduce to standard sine ratios; R1 use the fundamental sine limit; M1 evaluate the remaining factors; A1 result. Hybrid method: R1 identify the initial 0/0 form, A1 the first l’Hôpital quotient, A1 rearrange into normalized sine ratios, R1 invoke the fundamental sine limit, M1 evaluate the remaining limits, A1 obtain 3/2.