Key Concepts and Formulas

• Quadratic Equation Roots: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Limit of Trigonometric Functions: The standard limit $\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$ is crucial. This can be derived using L'Hopital's rule or trigonometric identities.
• Algebraic Manipulation: Factoring, simplifying expressions, and rationalizing denominators are essential for evaluating limits.

Step-by-Step Solution

Step 1: Find the roots of the given quadratic equation. The given equation is $2x^2 + x - 2 = 0$. Using the quadratic formula, the roots are: $x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-2)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 16}}{4} = \frac{-1 \pm \sqrt{17}}{4}$ Since $a > 0$ is a root, we have $a = \frac{-1 + \sqrt{17}}{4}$.

Step 2: Analyze the expression inside the cosine function and the denominator of the limit. The expression inside the cosine is $2 + x - 2x^2$. We can rewrite this by factoring out $-2$: $2 + x - 2x^2 = -2(x^2 - \frac{1}{2}x - 1)$ From the original quadratic equation $2x^2 + x - 2 = 0$, we know that if $a$ is a root, then $2a^2 + a - 2 = 0$. This implies $a^2 = 1 - \frac{a}{2}$. Also, if $a$ and $b$ are the roots of $2x^2 + x - 2 = 0$, then $a+b = -1/2$ and $ab = -1$. The roots are $a = \frac{-1 + \sqrt{17}}{4}$ and $b = \frac{-1 - \sqrt{17}}{4}$. Let's check the relation between the roots and $a$. We can write $2x^2 + x - 2 = 2(x-a)(x-b)$. Now consider the term $2 + x - 2x^2$. We can also write $2 + x - 2x^2 = -2x^2 + x + 2$. We can factor this expression. Since $a$ is a root of $2x^2 + x - 2 = 0$, then $2a^2 + a - 2 = 0$. Consider the expression $2 + x - 2x^2$. Let's evaluate it at $x = 1/a$. Since $2a^2 + a - 2 = 0$, dividing by $a^2$ (since $a \neq 0$), we get $2 + \frac{1}{a} - \frac{2}{a^2} = 0$, which means $\frac{2}{a^2} = 2 + \frac{1}{a}$. This doesn't directly help with $2 + x - 2x^2$. Let's try to express $2+x-2x^2$ in terms of $(x-1/a)$. Since $a$ is a root of $2x^2+x-2=0$, we have $2a^2+a-2=0$. Let's consider the expression $2+x-2x^2$. We are interested in the limit as $x \to 1/a$. If $x = 1/a$, then $2 + \frac{1}{a} - 2(\frac{1}{a})^2 = 2 + \frac{1}{a} - \frac{2}{a^2}$. From $2a^2+a-2=0$, we have $2 = 2a^2+a$. So, $2+x-2x^2 = (2a^2+a) + x - 2x^2$. This is not directly helpful. Let's try to factor $2+x-2x^2$. We know that $2x^2+x-2 = 0$ has roots $a$ and $b = \frac{-1-\sqrt{17}}{4}$. So, $2x^2+x-2 = 2(x-a)(x-b)$. Consider the expression $2+x-2x^2 = -(2x^2-x-2)$. This is not directly related to $2x^2+x-2$. Let's re-examine the problem statement and the current solution. The current solution uses $2+x-2x^2=-2\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)$. Let's verify this. If $x = 1/a$, then $2+1/a-2/a^2$. From $2a^2+a-2=0$, multiply by $1/a^2$: $2+1/a-2/a^2=0$. So, $x=1/a$ is a root of $2+x-2x^2=0$ if we substitute $x=1/a$. This is incorrect. Let's re-evaluate $2+x-2x^2$ as $x \to 1/a$. We are given that $a$ is a root of $2x^2+x-2=0$. So $2a^2+a-2=0$. Let's consider the expression $f(x) = 2+x-2x^2$. We can write $2+x-2x^2 = -2(x^2 - \frac{1}{2}x - 1)$. Let's try to express $2+x-2x^2$ in terms of $(x-1/a)$. Let $y = x - 1/a$. Then $x = y + 1/a$. $2 + (y+1/a) - 2(y+1/a)^2 = 2 + y + 1/a - 2(y^2 + 2y/a + 1/a^2)$ $= 2 + y + 1/a - 2y^2 - 4y/a - 2/a^2$ $= -2y^2 + y(1 - 4/a) + (2 + 1/a - 2/a^2)$. Since $2a^2+a-2=0$, we have $2+1/a-2/a^2 = 0$. So, $2+x-2x^2 = -2(x-1/a)^2 + (1-4/a)(x-1/a)$. This is a quadratic in $(x-1/a)$. Let's look at the denominator: $(1-ax)^2 = a^2(1/a - x)^2 = a^2(x-1/a)^2$.

Step 3: Use the limit of $(1-\cos \theta)/\theta^2$. The limit is of the form $\frac{1-\cos(\text{something})}{\text{something}^2}$. We need to manipulate the expression to use the standard limit $\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$. Let $u = 2+x-2x^2$. As $x \to 1/a$, $u \to 2 + 1/a - 2(1/a)^2$. From $2a^2+a-2=0$, divide by $a^2$: $2 + 1/a - 2/a^2 = 0$. So, as $x \to 1/a$, $u \to 0$. The limit becomes: $\lim _{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-a x)^2} = \lim _{x \rightarrow \frac{1}{a}} 16 \cdot \frac{1-\cos \left(2+x-2 x^2\right)}{\left(2+x-2 x^2\right)^2} \cdot \frac{\left(2+x-2 x^2\right)^2}{(1-a x)^2}$ The first part of the product tends to $16 \cdot \frac{1}{2} = 8$. Now we need to evaluate $\lim _{x \rightarrow \frac{1}{a}} \frac{\left(2+x-2 x^2\right)^2}{(1-a x)^2}$. We found that $2+x-2x^2 = -2(x-1/a)^2 + (1-4/a)(x-1/a)$. So, $(2+x-2x^2)^2 = \left[-2(x-1/a)^2 + (1-4/a)(x-1/a)\right]^2$ $= (x-1/a)^2 \left[-2(x-1/a) + (1-4/a)\right]^2$. The denominator is $(1-ax)^2 = a^2(x-1/a)^2$. So, the ratio is $\frac{(x-1/a)^2 \left[-2(x-1/a) + (1-4/a)\right]^2}{a^2(x-1/a)^2}$. As $x \to 1/a$, $(x-1/a) \to 0$. The limit of the ratio becomes $\lim _{x \rightarrow \frac{1}{a}} \frac{\left[-2(x-1/a) + (1-4/a)\right]^2}{a^2} = \frac{(1-4/a)^2}{a^2}$.

Step 4: Calculate the value of $1-4/a$. We have $a = \frac{\sqrt{17}-1}{4}$. So, $\frac{1}{a} = \frac{4}{\sqrt{17}-1} = \frac{4(\sqrt{17}+1)}{(\sqrt{17}-1)(\sqrt{17}+1)} = \frac{4(\sqrt{17}+1)}{17-1} = \frac{4(\sqrt{17}+1)}{16} = \frac{\sqrt{17}+1}{4}$. Then $1 - \frac{4}{a} = 1 - 4 \left(\frac{\sqrt{17}+1}{4}\right) = 1 - (\sqrt{17}+1) = -\sqrt{17}$. So, $(1-4/a)^2 = (-\sqrt{17})^2 = 17$.

Step 5: Calculate the value of $a^2$. Since $a$ is a root of $2x^2+x-2=0$, we have $2a^2+a-2=0$. $2a^2 = 2-a$. $a^2 = \frac{2-a}{2} = 1 - \frac{a}{2}$. Substitute $a = \frac{\sqrt{17}-1}{4}$: $a^2 = 1 - \frac{1}{2} \left(\frac{\sqrt{17}-1}{4}\right) = 1 - \frac{\sqrt{17}-1}{8} = \frac{8 - (\sqrt{17}-1)}{8} = \frac{9-\sqrt{17}}{8}$. Alternatively, $a^2 = \left(\frac{\sqrt{17}-1}{4}\right)^2 = \frac{(\sqrt{17})^2 - 2\sqrt{17} + 1^2}{16} = \frac{17 - 2\sqrt{17} + 1}{16} = \frac{18 - 2\sqrt{17}}{16} = \frac{9 - \sqrt{17}}{8}$.

Step 6: Combine the results to find the limit. The limit is $8 \cdot \frac{(1-4/a)^2}{a^2}$. Limit $= 8 \cdot \frac{17}{\frac{9-\sqrt{17}}{8}} = 8 \cdot 17 \cdot \frac{8}{9-\sqrt{17}} = \frac{64 \cdot 17}{9-\sqrt{17}}$. Rationalize the denominator: Limit $= \frac{64 \cdot 17}{9-\sqrt{17}} \cdot \frac{9+\sqrt{17}}{9+\sqrt{17}} = \frac{64 \cdot 17 (9+\sqrt{17})}{81 - 17} = \frac{64 \cdot 17 (9+\sqrt{17})}{64} = 17(9+\sqrt{17})$.

Step 7: Express the limit in the form $\alpha + \beta \sqrt{17}$ and find $\alpha + \beta$. The limit is $17(9+\sqrt{17}) = 153 + 17\sqrt{17}$. We are given that the limit is equal to $\alpha + \beta \sqrt{17}$, where $\alpha, \beta \in \mathbb{Z}$. Comparing $153 + 17\sqrt{17}$ with $\alpha + \beta \sqrt{17}$, we get $\alpha = 153$ and $\beta = 17$. Both $\alpha$ and $\beta$ are integers. We need to find $\alpha + \beta$. $\alpha + \beta = 153 + 17 = 170$.

Let's recheck the current solution's step where it says: $2+x-2 x^2=-2\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)$ Let's verify this factorization. If $x = 1/a$, then $2+1/a-2/a^2 = 0$. If $x = -\frac{4}{\sqrt{17}+1}$, then $2 - \frac{4}{\sqrt{17}+1} - 2 \left(\frac{-4}{\sqrt{17}+1}\right)^2 = 0$. $2 - \frac{4(\sqrt{17}-1)}{16} - 2 \frac{16}{(\sqrt{17}+1)^2} = 2 - \frac{\sqrt{17}-1}{4} - \frac{32}{17+1+2\sqrt{17}}$ $= 2 - \frac{\sqrt{17}-1}{4} - \frac{32}{18+2\sqrt{17}} = 2 - \frac{\sqrt{17}-1}{4} - \frac{16}{9+\sqrt{17}}$ $= 2 - \frac{\sqrt{17}-1}{4} - \frac{16(9-\sqrt{17})}{81-17} = 2 - \frac{\sqrt{17}-1}{4} - \frac{16(9-\sqrt{17})}{64}$ $= 2 - \frac{\sqrt{17}-1}{4} - \frac{9-\sqrt{17}}{4} = 2 - \frac{\sqrt{17}-1+9-\sqrt{17}}{4} = 2 - \frac{8}{4} = 2 - 2 = 0$. So, the roots of $2+x-2x^2=0$ are $1/a$ and $-4/(\sqrt{17}+1)$. Therefore, $2+x-2x^2 = -2(x-1/a)(x - (-4/(\sqrt{17}+1)))$. The factorization in the current solution is correct.

Now, let's re-evaluate the limit using this factorization. \lim _\limits{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-a x)^2} Let $u = 2+x-2x^2$. As $x \to 1/a$, $u \to 0$. The limit is $\lim _{u \rightarrow 0} \frac{16(1-\cos u)}{u^2} \cdot \lim _{x \rightarrow \frac{1}{a}} \frac{u^2}{(1-ax)^2}$. The first part is $16 \cdot \frac{1}{2} = 8$. The second part is $\lim _{x \rightarrow \frac{1}{a}} \frac{(-2(x-1/a)(x+4/(\sqrt{17}+1)))^2}{(1-ax)^2}$ $= \lim _{x \rightarrow \frac{1}{a}} \frac{4(x-1/a)^2 (x+4/(\sqrt{17}+1))^2}{a^2(x-1/a)^2}$ $= \lim _{x \rightarrow \frac{1}{a}} \frac{4 (x+4/(\sqrt{17}+1))^2}{a^2}$ Substitute $x=1/a$: $= \frac{4 (1/a+4/(\sqrt{17}+1))^2}{a^2}$. We know $\frac{1}{a} = \frac{\sqrt{17}+1}{4}$. So, $\frac{1}{a} + \frac{4}{\sqrt{17}+1} = \frac{\sqrt{17}+1}{4} + \frac{4(\sqrt{17}-1)}{16} = \frac{\sqrt{17}+1}{4} + \frac{\sqrt{17}-1}{4} = \frac{2\sqrt{17}}{4} = \frac{\sqrt{17}}{2}$. The limit of the second part is $\frac{4 (\sqrt{17}/2)^2}{a^2} = \frac{4(17/4)}{a^2} = \frac{17}{a^2}$. The total limit is $8 \cdot \frac{17}{a^2} = \frac{136}{a^2}$. We found $a^2 = \frac{9-\sqrt{17}}{8}$. Limit $= \frac{136}{(9-\sqrt{17})/8} = \frac{136 \cdot 8}{9-\sqrt{17}} = \frac{1088}{9-\sqrt{17}}$. Rationalize: $\frac{1088(9+\sqrt{17})}{81-17} = \frac{1088(9+\sqrt{17})}{64} = 17(9+\sqrt{17}) = 153 + 17\sqrt{17}$. This matches our previous result.

Let's re-examine the current solution's calculation from step 3 onwards: \lim _\limits{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-a x)^2} $=\lim _{x \rightarrow \frac{1}{a}} \frac{32 \sin ^2\left(\frac{2+x-2 x^2}{2}\right)}{a^2\left(\frac{1}{a}-x\right)^2}$ This step uses $1-\cos \theta = 2\sin^2(\theta/2)$ and $(1-ax)^2 = a^2(1/a-x)^2$. This is correct. $=\lim _{x \rightarrow \frac{1}{a}} \frac{\left(x+\frac{4}{\sqrt{17}+1}\right)^2 32 \cdot\left(\sin \left(\frac{1}{2} \cdot(-2)\right)\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2}{a^2\left(\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2}$ This step seems to have an error in the substitution of $2+x-2x^2$. The factorization is $2+x-2x^2 = -2(x-1/a)(x+4/(\sqrt{17}+1))$. So, $\frac{2+x-2x^2}{2} = -(x-1/a)(x+4/(\sqrt{17}+1))$. The term inside the sine squared should be this. The current solution has $\sin \left(\frac{1}{2} \cdot(-2)\right)\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)$. This implies $\sin(-1) \cdot (\dots)$. This is incorrect. The argument of sine should be $\frac{2+x-2x^2}{2}$. So, $\sin^2\left(\frac{2+x-2x^2}{2}\right) = \sin^2\left(\frac{-2(x-1/a)(x+4/(\sqrt{17}+1))}{2}\right) = \sin^2\left(-(x-1/a)(x+4/(\sqrt{17}+1))\right)$. The limit expression should be: $\lim _{x \rightarrow \frac{1}{a}} \frac{32 \sin^2\left(-(x-1/a)(x+4/(\sqrt{17}+1))\right)}{a^2\left(\frac{1}{a}-x\right)^2}$ $=\lim _{x \rightarrow \frac{1}{a}} \frac{32 \sin^2\left((x-1/a)(x+4/(\sqrt{17}+1))\right)}{a^2\left(x-\frac{1}{a}\right)^2}$ $=\lim _{x \rightarrow \frac{1}{a}} \frac{32 \sin^2\left((x-1/a)(x+4/(\sqrt{17}+1))\right)}{(x-1/a)^2} \cdot \frac{1}{a^2}$ Using $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, so $\lim_{\theta \to 0} \frac{\sin^2 \theta}{\theta^2} = 1$. Let $\theta = (x-1/a)(x+4/(\sqrt{17}+1))$. As $x \to 1/a$, $\theta \to 0$. We need to divide by $\theta^2 = (x-1/a)^2(x+4/(\sqrt{17}+1))^2$. So, $\lim _{x \rightarrow \frac{1}{a}} \frac{32 \sin^2\left((x-1/a)(x+4/(\sqrt{17}+1))\right)}{(x-1/a)^2(x+4/(\sqrt{17}+1))^2} \cdot \frac{(x-1/a)^2(x+4/(\sqrt{17}+1))^2}{(x-1/a)^2} \cdot \frac{1}{a^2}$ $= 32 \cdot 1 \cdot \lim _{x \rightarrow \frac{1}{a}} (x+4/(\sqrt{17}+1))^2 \cdot \frac{1}{a^2}$ Substitute $x=1/a$: $= 32 \cdot (1/a+4/(\sqrt{17}+1))^2 \cdot \frac{1}{a^2}$ As calculated before, $1/a+4/(\sqrt{17}+1) = \sqrt{17}/2$. So, the limit is $32 \cdot (\sqrt{17}/2)^2 \cdot \frac{1}{a^2} = 32 \cdot (17/4) \cdot \frac{1}{a^2} = 8 \cdot 17 \cdot \frac{1}{a^2} = \frac{136}{a^2}$. This matches our earlier calculation.

Let's analyze the current solution's subsequent steps: $=2 \cdot\left(\frac{1}{a}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2$ This step has a factor of 2 instead of 32 and the term $(4/(\sqrt{17}-1))^2$. Let's check if $4/(\sqrt{17}-1)$ relates to anything. $4/(\sqrt{17}-1) = 4(\sqrt{17}+1)/16 = (\sqrt{17}+1)/4 = 1/a$. So this term is $(1/a)^2$. The current solution has a factor of 2, and then $(\frac{1}{a}+\frac{4}{\sqrt{17}+1})^2 \cdot (\frac{1}{a})^2$. This means it's $2 \cdot (\sqrt{17}/2)^2 \cdot (1/a)^2 = 2 \cdot (17/4) \cdot (1/a)^2 = (17/2) \cdot (1/a)^2$. This does not match $136/a^2$.

There seems to be a significant error in the current solution's algebraic manipulation from step 3 onwards.

Let's reconfirm the steps for clarity. $a = (\sqrt{17}-1)/4$. $1/a = (\sqrt{17}+1)/4$. $2+x-2x^2 = -2(x-1/a)(x+4/(\sqrt{17}+1))$. Denominator is $(1-ax)^2 = a^2(x-1/a)^2$. Limit is $\lim_{x \to 1/a} \frac{16(1-\cos(2+x-2x^2))}{(1-ax)^2}$. Let $y = 2+x-2x^2$. As $x \to 1/a$, $y \to 0$. $\lim_{x \to 1/a} \frac{16(1-\cos y)}{y^2} \frac{y^2}{(1-ax)^2} = 16 \cdot \frac{1}{2} \cdot \lim_{x \to 1/a} \frac{(-2(x-1/a)(x+4/(\sqrt{17}+1)))^2}{a^2(x-1/a)^2}$ $= 8 \cdot \lim_{x \to 1/a} \frac{4(x-1/a)^2 (x+4/(\sqrt{17}+1))^2}{a^2(x-1/a)^2}$ $= 8 \cdot \frac{4}{a^2} \lim_{x \to 1/a} (x+4/(\sqrt{17}+1))^2$ $= \frac{32}{a^2} (1/a+4/(\sqrt{17}+1))^2$ $= \frac{32}{a^2} (\sqrt{17}/2)^2 = \frac{32}{a^2} \frac{17}{4} = \frac{8 \cdot 17}{a^2} = \frac{136}{a^2}$. $a^2 = (\frac{\sqrt{17}-1}{4})^2 = \frac{18-2\sqrt{17}}{16} = \frac{9-\sqrt{17}}{8}$. Limit $= \frac{136}{(9-\sqrt{17})/8} = \frac{136 \cdot 8}{9-\sqrt{17}} = \frac{1088}{9-\sqrt{17}}$. $= \frac{1088(9+\sqrt{17})}{81-17} = \frac{1088(9+\sqrt{17})}{64} = 17(9+\sqrt{17}) = 153 + 17\sqrt{17}$. $\alpha = 153$, $\beta = 17$. $\alpha + \beta = 153 + 17 = 170$.

The current solution's final calculation steps are incorrect, but the final answer $\alpha+\beta=170$ is correct. This suggests that the correct answer might have been obtained through a different path or a lucky mistake in calculation that led to the right final number.

Let's trace the calculation in the provided solution again: $=2 \cdot\left(\frac{1}{a}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2$ We know $\frac{1}{a} = \frac{\sqrt{17}+1}{4}$ and $\frac{4}{\sqrt{17}-1} = \frac{4(\sqrt{17}+1)}{16} = \frac{\sqrt{17}+1}{4} = \frac{1}{a}$. So the expression is $2 \cdot (\frac{\sqrt{17}}{2})^2 \cdot (\frac{1}{a})^2 = 2 \cdot \frac{17}{4} \cdot \frac{1}{a^2} = \frac{17}{2a^2}$. The next step is: $= \frac{17 \times 4}{18-2 \sqrt{17}}=\frac{68}{9-\sqrt{17}}$ This means $\frac{17}{2a^2} = \frac{17 \times 4}{18-2 \sqrt{17}}$. $\frac{1}{2a^2} = \frac{4}{18-2\sqrt{17}} = \frac{2}{9-\sqrt{17}}$. $9-\sqrt{17} = 4a^2$. We know $a^2 = \frac{9-\sqrt{17}}{8}$. So $4a^2 = 4 \frac{9-\sqrt{17}}{8} = \frac{9-\sqrt{17}}{2}$. This means $9-\sqrt{17} = \frac{9-\sqrt{17}}{2}$, which implies $9-\sqrt{17}=0$, which is false. There is a clear inconsistency in the provided solution's steps.

Let's assume the current solution meant to calculate $\frac{17}{2a^2}$ and then proceed. $\frac{17}{2a^2} = \frac{17}{2 \cdot \frac{9-\sqrt{17}}{8}} = \frac{17}{\frac{9-\sqrt{17}}{4}} = \frac{17 \cdot 4}{9-\sqrt{17}} = \frac{68}{9-\sqrt{17}}$. This step seems to be derived from $\frac{17}{2a^2}$. The next step: $=17(9+\sqrt{17})$ This means $\frac{68}{9-\sqrt{17}} = 17(9+\sqrt{17})$. $\frac{4}{9-\sqrt{17}} = 9+\sqrt{17}$. $4 = (9-\sqrt{17})(9+\sqrt{17}) = 81-17 = 64$. $4 = 64$, which is false.

The provided solution has significant errors in its derivation. However, the final answer $\alpha+\beta=170$ is correct. My derivation leads to $\alpha=153, \beta=17$, so $\alpha+\beta=170$.

Common Mistakes & Tips

• Factorization Errors: Be very careful when factoring expressions, especially when dealing with negative signs and fractions. Double-check your factorization by expanding it.
• Algebraic Simplification: Errors in rationalizing denominators or simplifying fractions can lead to incorrect final answers.
• Standard Limit Application: Ensure that the argument of the cosine function in the numerator tends to zero as $x$ approaches the limit point. If not, use Taylor series expansion or L'Hopital's rule.
• Correct Root Identification: Make sure to use the correct root ($a>0$) when substituting values.

Summary

The problem requires evaluating a limit involving a trigonometric function and a quadratic expression. The key steps involve finding the roots of the given quadratic equation, manipulating the expression inside the cosine function and the denominator to apply the standard limit $\lim_{\theta \to 0} \frac{1-\cos \theta}{\theta^2} = \frac{1}{2}$, and performing careful algebraic simplification. By correctly applying these concepts, we found the limit to be $153 + 17\sqrt{17}$, which gives $\alpha = 153$ and $\beta = 17$. Therefore, $\alpha + \beta = 170$.

The final answer is \boxed{170}.