Problem Analysis #17

[THOUGHT]

Given limit problem involves two main components in the integrand:

1. $\frac{a}{2} \int_0^x \frac{1}{1-t^2}\ dt$ which is an improper integral that can be solved using logarithmic rules,
2. $bx\cos x$, a product of trigonometric and polynomial functions.

First, we evaluate the integral in the limit expression: $\int_0^x \frac{1}{1-t^2} = \left[ \frac{1}{2} \ln|1+t| - \frac{1}{4} \ln|1-t^2| \right]_0^x$ $= \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right) - \frac{1}{4} \ln(1-x^2)$.

Next, we apply L'Hopital's Rule to the limit as $x \rightarrow 0$, because $\lim_{x \to 0} (1-x)^{-\frac{1}{2}}$ and $\lim_{x \to 0} (1-x^2)^{-\frac{1}{4}}$ are both indeterminate forms of type $\infty / \infty$.

Applying L'Hopital's Rule: $\lim_{x \rightarrow 0} \frac{\left(\frac{a}{2}\right)\left[\frac{1/x}{(1+x)/(1-x)} - \frac{-x/2(1-x^2)^{-\frac{3}{4}}}{-2x(1-t^2)^{-\frac{1}{2}}}\right] + bx(\cos x - \sin x)}{3x^2}$ $= \lim_{x \rightarrow 0} \left[\left(\frac{a}{2}\right)\left(\frac{1}{1-x^2} - \frac{x(1+x)}{4(1-x^3)^{\frac{3}{4}}}\right) + b(x\cos x - \sin x)\right]$ $= \lim_{x \rightarrow 0} \left[\left(\frac{a}{2}\right)\left(\frac{1}{1-x^2} - \frac{x(1+x)}{4(1-x)^{\frac{3}{4}}}\right) + bx\left(\cos x - \frac{\sin x}{x}\right) - b\sin x\right]$ $= \lim_{x \rightarrow 0} \left[\left(\frac{a}{2}\right)\left(1 + \frac{x^2}{2} + O(x^3)\right) + bx\left(1 - \frac{1}{x} + O(x^2)\right) - b\sin x\right]$ $= \lim_{x \rightarrow 0} \left[\frac{a}{2} + \frac{ax^2}{4} + bx - \frac{b}{x} + O(x^3)\right]$ $= \lim_{x \rightarrow 0} \left[a + \frac{1}{2}(a+2b)x - b\right] = a - b$.

Given limit equals $2$, we have: $a - b = 2$.

Now, to find $(a+b)^2$, we square both sides of the equation and then add $2ab$: $(a-b)^2 + 2ab = 4 + 2ab$ $(a+b)^2 = 4 + 2ab$

But $a+b=2$ (from $a-b=2$), so: $2^2 = 4 + 2(a+b)$ $4 = 4 + 4 = 8$

Hence, $(a+b)^2 = 8$, and $100(a+b)^2 = 800$.

$\boxed{800}$