Find the positive root $a$ of $2x^2+x-2=0$.

Solve: $2x^2+x-2=0$ Using the quadratic formula, $x=\frac{-1\pm\sqrt{1+16}}{4}=\frac{-1\pm\sqrt{17}}{4}$ Since $a>0$, $a=\frac{-1+\sqrt{17}}{4}$

Evaluate the limit $L=\lim_{x\to 1/a}\frac{16\left(1-\cos(2+x-2x^2)\right)}{(1-ax)^2}$

Let $u=2+x-2x^2$ At $x=\frac1a$, $u=2+\frac1a-2\left(\frac1a\right)^2$

Since $a$ is a root of $2a^2+a-2=0$, dividing by $a^2$ gives $2+\frac1a-\frac{2}{a^2}=0$ Hence at $x=\frac1a$, $u\to 0$

Therefore we can use $1-\cos u \sim \frac{u^2}{2} \quad (u\to 0)$ So

Factor $2+x-2x^2$.

Since $x=1/a$ is a root of $2+x-2x^2=0$, write $2+x-2x^2=-2\left(x-\frac1a\right)(x-r)$ But it is easier to differentiate or factor via the other root.

Let $f(x)=2+x-2x^2$ Then $f'(x)=1-4x$ Since $f(1/a)=0$ and denominator also vanishes linearly, $\lim_{x\to 1/a}\frac{f(x)}{1-ax}=\frac{f'(1/a)}{-a}$ by L'Hospital's Rule.

So,

Now simplify using $2a^2+a-2=0$.

From $a=\frac{-1+\sqrt{17}}{4}$ we compute directly: $4-a=4-\frac{-1+\sqrt{17}}{4}=\frac{17-\sqrt{17}}{4}$ and $a^2=\left(\frac{-1+\sqrt{17}}{4}\right)^2=\frac{9-\sqrt{17}}{8}$

Therefore

Thus, $\alpha=153,\quad \beta=17$ Therefore, $\alpha+\beta=170$

Comparison with stored answer:

Stored correct answer = $170$.

Our derived answer matches it.