#35 Measure Your Calibre

Write it(L) as: $L= \beta \cdot \displaystyle{\lim_{x \rightarrow 0} \dfrac{x^3 \times \overbrace{\color{#D61F06}{\frac{\sin(\beta x)}{\beta x}}}^1}{\alpha x - \sin x}}~~\left(\small{\frac 00 }\right)$ Apply Lhopital rule: $L=\beta\cdot \displaystyle{\lim_{x \rightarrow 0} \dfrac{3x^2 }{\alpha - \cos x}}$ Since numerator is tending to zero, for limit to be $1(\ne 0)$ , denominator must also tend to zero i.e $\color{#20A900}{\alpha=1}$ . Then :

$L=3\beta\cdot\displaystyle{\lim_{x \rightarrow 0} \dfrac{1}{\underbrace{\color{#D61F06}{\dfrac{1 - \cos x}{x^2}}}_{0.5}}}=6\beta=1$ $\implies \color{#20A900}{\beta=\dfrac 16}$

$\therefore 6(\alpha+\beta)+2=\boxed{9}$

Using Taylor expansion : $\displaystyle \sin x=\sum\limits_{n=0}^{\infty}\dfrac{x^{2n+1}(-1)^n}{(2n+1)!}$

$\displaystyle \begin{aligned} L&=\lim_{x\to 0}\dfrac{x^2\sin \beta x}{\alpha x-\sin x} \\&=\lim_{x\to 0} \dfrac{x^2\left(\beta x-\dfrac{\beta^3 x^3}{3!}+\cdots\right)}{\alpha x-\left(x-\dfrac{x^3}{3!}+\cdots\right)} \\ &= \lim_{x\to 0}\dfrac{\beta-\dfrac{\beta^3 x^2}{3!}+\cdots}{\dfrac{\alpha-1}{x^2}+\dfrac{1}{3!}+\cdots} \\ \end{aligned}$

This forces $\displaystyle \alpha=1$ ,

$\displaystyle \begin{aligned} L&=\lim_{x\to 0} \dfrac{\beta}{\dfrac{1}{3!}}=1\end{aligned}$

Thus $\beta=\dfrac{1}{6}$ making the answer $\boxed{6(\alpha+\beta)+2=9}$

Relevant wiki: Maclaurin Series

$\begin{aligned} L & = \lim_{x \to 0} \frac {x^2 \color{#3D99F6} \sin (\beta x)}{\alpha x - \color{#3D99F6} \sin x} & \small \color{#3D99F6} \text{Using Maclaurin series} \\ & = \lim_{x \to 0} \frac {x^2 \color{#3D99F6} \left(\beta x - \frac {(\beta x)^3}{3!} + \frac {(\beta x)^5}{5!} - \frac {(\beta x)^7}{7!} + ... \right)}{\alpha x - \color{#3D99F6} \left(x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + ... \right)} \\ & = \lim_{x \to 0} \frac {\beta x^3 - \frac {\beta^3 x^5}{3!} + \frac {\beta^5 x^7}{5!} - \frac {\beta^7 x^9}{7!} + ...}{{\color{#3D99F6}\alpha} x - x + \frac {x^3}{3!} - \frac {x^5}{5!} + \frac {x^7}{7!} + ...} & \small \color{#3D99F6} \text{Putting }\alpha = 1 \\ & = \lim_{x \to 0} \frac {\beta x^3 - \frac {\beta^3 x^5}{3!} + \frac {\beta^5 x^7}{5!} - \frac {\beta^7 x^9}{7!} + ...}{0+ \frac {x^3}{3!} - \frac {x^5}{5!} + \frac {x^7}{7!} + ...} & \small \color{#3D99F6} \text{Dividing up and down by }x^3 \\ & = \lim_{x \to 0} \frac {\beta - \frac {\beta^3 x^2}{3!} + \frac {\beta^5 x^4}{5!} - \frac {\beta^7 x^6}{7!} + ...}{\frac 1{3!} - \frac {x^2}{5!} + \frac {x^4}{7!} + ...} \\ & = 6 \beta \end{aligned}$

Since $L = 1$ , $\implies \beta = \dfrac 16$ , $\implies 6(\alpha + \beta) + 2 = 6\left(1+\dfrac 16 \right) + 2 = \boxed{9}$