The problem asks for the value of $a^2 + b^2 + c^2$ given a limit equation involving these constants.

Limit Analysis

Let the limit be L. We have:

$ L = \lim_{x \to 0} \frac{e^{(a-1)x} + 2 \cos bx + (c-2)e^{-x}}{x \cos x - \log_e(1+x)} = 2 $

First, evaluate the denominator at $x=0$: $D(0) = 0 \cdot \cos(0) - \log_e(1+0) = 0 \cdot 1 - \log_e(1) = 0 - 0 = 0$. Since the limit is finite and non-zero, the numerator must also be zero at $x=0$.

Evaluate the numerator at $x=0$: $N(0) = e^{(a-1)0} + 2 \cos(b \cdot 0) + (c-2)e^{-0} = e^0 + 2 \cos(0) + (c-2)e^0 = 1 + 2(1) + (c-2)(1) = 1 + 2 + c - 2 = c+1$.

Setting $N(0)=0$, we get $c+1 = 0$, which implies $c = -1$.

Applying Taylor Series Expansion

Now substitute $c=-1$ into the numerator expression:

$N(x) = e^{(a-1)x} + 2 \cos bx + (-1-2)e^{-x} = e^{(a-1)x} + 2 \cos bx - 3e^{-x}$

We use the Taylor expansions around $x=0$:

• $e^u = 1 + u + \frac{u^2}{2!} + ...$
• $\cos u = 1 - \frac{u^2}{2!} + ...$
• $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ...$

Expand the denominator $D(x)$:

$D(x) = x \cos x - \log_e(1+x)$

$D(x) = x(1 - \frac{x^2}{2!} + ...) - (x - \frac{x^2}{2} + \frac{x^3}{3} - ...)$

$D(x) = (x - \frac{x^3}{2} + ...) - (x - \frac{x^2}{2} + \frac{x^3}{3} - ...)$

$D(x) = \frac{x^2}{2} - (\frac{1}{2} + \frac{1}{3})x^3 + ... = \frac{x^2}{2} - \frac{5x^3}{6} + ...$

The lowest power term in the denominator is $\frac{x^2}{2}$.

Expand the numerator $N(x)$ using $c=-1$:

$N(x) = (1 + (a-1)x + \frac{(a-1)^2 x^2}{2} + ...) + 2(1 - \frac{(bx)^2}{2} + ...) - 3(1 - x + \frac{x^2}{2} - ...)$

Group terms by powers of $x$:

Constant term: $1 + 2 - 3 = 0$. (Consistent with $c=-1$)

$x$ term: $(a-1)x + 3x = (a-1+3)x = (a+2)x$.

For the limit to be finite, the coefficient of the $x$ term in the numerator must be zero (since the denominator's lowest power is $x^2$).

$a+2 = 0 \implies a = -2$.

$x^2$ term: $\frac{(a-1)^2 x^2}{2} - \frac{2b^2 x^2}{2} - \frac{3x^2}{2} = (\frac{(a-1)^2}{2} - b^2 - \frac{3}{2})x^2$.

Substitute $a=-2$: $(\frac{(-2-1)^2}{2} - b^2 - \frac{3}{2})x^2 = (\frac{(-3)^2}{2} - b^2 - \frac{3}{2})x^2 = (\frac{9}{2} - b^2 - \frac{3}{2})x^2 = (\frac{6}{2} - b^2)x^2 = (3 - b^2)x^2$.

So, the numerator is approximately $(3 - b^2)x^2$ for small $x$.

Calculating the Limit Value

The limit becomes:

$ L = \lim_{x \to 0} \frac{(3 - b^2)x^2 + ...}{\frac{x^2}{2} + ...} = \frac{3 - b^2}{1/2} = 2(3 - b^2) $

We are given $L=2$:

$2(3 - b^2) = 2$

$3 - b^2 = 1$

$b^2 = 2$.

Final Calculation

We have found the values:

• $a = -2 \implies a^2 = (-2)^2 = 4$
• $b^2 = 2$
• $c = -1 \implies c^2 = (-1)^2 = 1$

Therefore, $a^2 + b^2 + c^2 = 4 + 2 + 1 = 7$.