If lim_{x rightarrow 0} frac{e^{(a-1)x} + 2 c | vedclass

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(C) Using the Taylor series expansion for $x \rightarrow 0$: $e^{(a-1)x} = 1 + (a-1)x + \frac{(a-1)^2 x^2}{2} + O(x^3)$ $2 \cos(bx) = 2(1 - \frac{b^2

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$7$

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(C) Using the Taylor series expansion for $x ightarrow 0$: $e^{(a-1)x} = 1 + (a-1)x + \frac{(a-1)^2 x^2}{2} + O(x^3)$ $2 \cos(bx) = 2(1 - \frac{b^2 x^2}{2} + O(x^4)) = 2 - b^2 x^2 + O(x^4)$ $(c-2)e^{-x} = (c-2)(1 - x + \frac{x^2}{2} + O(x^3))$ Denominator: $x(1 - \frac{x^2}{2}) - (x - \frac{x^2}{2} + \frac{x^3}{3}) = x - \frac{x^3}{2} - x + \frac{x^2}{2} - \frac{x^3}{3} = \frac{x^2}{2} - \frac{5x^3}{6} + O(x^4)$ For the limit to exist and equal $2$, the coefficients of $x^0$ and $x^1$ in the numerator must be $0$: Constant term: $1 + 2 + c - 2 = c + 1 = 0 \Rightarrow c = -1$ Coefficient of $x$: $(a-1) - (c-2) = a - 1 - c + 2 = a - c + 1 = 0$ $\Rightarrow a - (-1) + 1 = 0$ $\Rightarrow a = -2$ Coefficient of $x^2$: $\frac{(a-1)^2}{2} - b^2 + \frac{c-2}{2} = 2 	imes \frac{1}{2} = 1$ $\frac{(-2-1)^2}{2} - b^2 + \frac{-1-2}{2} = 1$ $\Rightarrow \frac{9}{2} - b^2 - \frac{3}{2} = 1$ $\Rightarrow 3 - b^2 = 1$ $\Rightarrow b^2 = 2$ Thus, $a^2 + b^2 + c^2 = (-2)^2 + 2 + (-1)^2 = 4 + 2 + 1 = 7$.

If $\lim_{x \rightarrow 0} \frac{e^{(a-1)x} + 2 \cos(bx) + (c-2)e^{-x}}{x \cos x - \log_{e}(1+x)} = 2$, then $a^{2} + b^{2} + c^{2}$ is equal to:

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