A calculus problem by Hamza A

$\Large \color{#302B94}{\boxed{\color{#D61F06}{\mathcal{METHOD}~1}}}$ Use expansion series of sin x i.e $\small{\color{#3D99F6}{\sin x =x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}\cdots}}$ so that limit simplifies to : $\displaystyle\lim _{ x\rightarrow 0 }\dfrac{1}{3!}-\dfrac{x^2}{5!}+\dfrac{x^4}{7!}\cdots$ $\Large=\dfrac{1}{3!}$ $\Large\color{#302B94}{\boxed{\color{#D61F06}{\mathcal{METHOD}~2}}}$ Use L'Hôpital's Rule and $\small{\color{#3D99F6}{\lim _{ x\rightarrow 0 }\dfrac{1-\cos x}{x^2}=\dfrac{1}{2}}}$ $\displaystyle\lim _{ x\rightarrow 0 }{ \frac { x-\sin { x } }{ { x }^{ 3 } } } =\displaystyle\lim _{ x\rightarrow 0 }{ \frac { 1-\cos { x } }{ { 3x }^{ 2} } }$ $\Large =\dfrac{1}{6}=\dfrac{1}{3!}$

$\huge \therefore 1+3=\color{#0C6AC7}{\boxed{{\color{forestgreen}{4}}}}$

Evaluating this we obtaun the answer $\Large \frac {1}{6}$ .
And, $6=3×2×1=3!$
So, $A$ = $1$
And, $B$ = $3$ .
So, $A+B$ = $1+3=4$