A calculus problem by Md Zuhair

Solution 1: by Maclaurin series

$\begin{aligned} L & = \lim_{x \to 0} \frac {1- \color{#3D99F6} \cos 2x}{1- \color{#3D99F6} \cos 3x} & \small \color{#3D99F6} \text{By Maclaurin series expansion} \\ & = \lim_{x \to 0} \frac {1-\left(1 - \frac {2^2x^2}{2!} + \frac {2^4x^4}{4!} - ...\right)}{1-\left(1 - \frac {3^2x^2}{2!} + \frac {3^4x^4}{4!} - ...\right)} \\ & = \lim_{x \to 0} \frac {\frac {2^2x^2}{2!} - \frac {2^4x^4}{4!} + \frac {2^6x^6}{6!} - ...}{\frac {3^2x^2}{2!} - \frac {3^4x^4}{4!} + \frac {3^6x^6}{6!} -...} & \small \color{#3D99F6} \text{Divide up and down by }x^2 \\ & = \lim_{x \to 0} \frac {\frac {2^2}{2!} - \frac {2^4x^2}{4!} + \frac {2^6x^4}{6!} - ...}{\frac {3^2}{2!} - \frac {3^4x^2}{4!} + \frac {3^6x^4}{6!} -...} \\ & = \frac {2^2}{3^2} = \frac 49 \end{aligned}$

$\implies a + b = 4+9 = \boxed{13}$

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Solution 2: by L'Hôpital's rule

$\begin{aligned} L & = \lim_{x \to 0} \frac {1- \cos 2x}{1- \cos 3x} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \frac {2 \sin 2x}{3 \sin 3x} & \small \color{#3D99F6} \text{A 0/0 case and apply L'Hôpital's rule again.} \\ & = \lim_{x \to 0} \frac {-4 \cos 2x}{-9 \cos 3x} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t }x \\ & = \frac 49 \end{aligned}$

$\implies a + b = 4+9 = \boxed{13}$

We use L-Hopital to solve this type of problem {(-)(sin2x) 2}/{(-)(sin3x) 3}=4/9 a=4 b=9 so a+b=13