Limits, Continuity and Differentiability Part 3

Concept:Use series expansions of $e^{(a-1)x}$, $\cos(bx)$, $e^{-x}$, $\cos x$ and $\log_e(1+x)$ near $x=0$, then equate coefficients to satisfy the given limit $2$.Explanation:Write the numerator expansion:$e^{(a-1)x} = 1 + (a-1)x + \dfrac{(a-1)^2}{2}x^2 + \cdots$$2\cos(bx) = 2\left(1 - \dfrac{b^2}{2}x^2 + \cdots\right)$$(c-2)e^{-x} = (c-2)\left(1 - x + \dfrac{x^2}{2} - \cdots\right)$Add them: constant term = $1 + 2 + (c-2) = c+1$Coefficient of $x$: $(a-1) + 0 + (c-2)(-1) = a - c + 1$Coefficient of $x^2$: $\dfrac{(a-1)^2}{2} - b^2 + \dfrac{c-2}{2}$Denominator: $x\cos x = x\left(1 - \dfrac{x^2}{2} + \cdots\right) = x - \dfrac{x^3}{2} + \cdots$$\log_e(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$Subtract: $x\cos x - \log_e(1+x) = \left(x - \dfrac{x^3}{2}\right) - \left(x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots\right) = \dfrac{x^2}{2} - \dfrac{5x^3}{6} + \cdots$Lowest power in denominator is $x^2$.For limit to be finite $2$, the constant and $x$ terms in numerator must vanish:$c+1 = 0 \Rightarrow c = -1$$a - c + 1 = 0 \Rightarrow a - (-1) + 1 = 0 \Rightarrow a = -2$Now the leading term of numerator is $\left(\dfrac{(a-1)^2}{2} - b^2 + \dfrac{c-2}{2}\right)x^2 = \left(\dfrac{(-3)^2}{2} - b^2 + \dfrac{-1-2}{2}\right)x^2 = \left(\dfrac{9}{2} - b^2 - \dfrac{3}{2}\right)x^2 = (3 - b^2)x^2$Denominator leading term is $\dfrac{x^2}{2}$.Thus $\lim\limits_{x\to0} \dfrac{(3-b^2)x^2}{x^2/2} = 2(3-b^2) = 2 \Rightarrow 3-b^2 = 1 \Rightarrow b^2 = 2$.Now $a^2 = (-2)^2 = 4$, $c^2 = (-1)^2 = 1$, so $a^2+b^2+c^2 = 4+2+1 = 7$.Answer:$7$ (Option D)