If $$\lim_{x \to 0} \frac{ax^2e^x - b\log_e(1+x) + cxe^{-x}}{x^2\sin x} = 1$$, then $$16(a^2 + b^2 + c^2)$$ is equal to

We need to find the value of $$16(a^2 + b^2 + c^2)$$ given that the following limit equals 1:

$$\lim_{x \to 0} \frac{ax^2 e^x - b\ln(1+x) + cxe^{-x}}{x^2 \sin x} = 1$$

We first expand each function using Taylor series around $$x = 0$$ up to $$x^3$$ (since the denominator behaves as $$\sim x^3$$).

$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$

$$e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots$$

$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$$

$$\sin x = x - \frac{x^3}{6} + \cdots$$

The denominator expands as

$$x^2 \sin x = x^2\Bigl(x - \frac{x^3}{6} + \cdots\Bigr) = x^3 - \frac{x^5}{6} + \cdots$$

As $$x \to 0$$, we truncate at $$x^3$$ since higher-order terms vanish relative to $$x^3$$.

Next, we expand the numerator term by term.

Term 1: $$ax^2 e^x = ax^2\Bigl(1 + x + \frac{x^2}{2} + \cdots\Bigr) = ax^2 + ax^3 + \cdots$$

Term 2: $$-b\ln(1+x) = -b\Bigl(x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\Bigr) = -bx + \frac{b}{2}x^2 - \frac{b}{3}x^3 + \cdots$$

Term 3: $$cxe^{-x} = cx\Bigl(1 - x + \frac{x^2}{2} - \cdots\Bigr) = cx - cx^2 + \frac{c}{2}x^3 + \cdots$$

Combining these, the numerator has the coefficients

of $$x^1$$: $$-b + c$$

of $$x^2$$: $$a + \frac{b}{2} - c$$

of $$x^3$$: $$a - \frac{b}{3} + \frac{c}{2}$$

For the limit to exist and equal 1, the numerator must be of order $$x^3$$. Therefore the coefficients of $$x$$ and $$x^2$$ must vanish, and the coefficient of $$x^3$$ divided by the coefficient 1 from the denominator must equal 1.

From the coefficient of $$x$$: $$-b + c = 0 \implies c = b$$

From the coefficient of $$x^2$$: $$a + \frac{b}{2} - c = 0$$ and since $$c = b$$, we get $$a = \frac{b}{2}$$

From the coefficient of $$x^3$$: $$a - \frac{b}{3} + \frac{c}{2} = 1$$. Substituting $$a = \tfrac{b}{2}$$ and $$c = b$$ yields

$$\frac{b}{2} - \frac{b}{3} + \frac{b}{2} = 1 \implies b - \frac{b}{3} = 1 \implies \frac{2b}{3} = 1 \implies b = \frac{3}{2}$$

Hence $$b = \frac{3}{2}$$, $$c = \frac{3}{2}$$, and $$a = \frac{3}{4}$$.

It follows that

$$a^2 + b^2 + c^2 = \frac{9}{16} + \frac{9}{4} + \frac{9}{4} = \frac{9}{16} + \frac{36}{16} + \frac{36}{16} = \frac{81}{16}$$

and therefore

$$16(a^2 + b^2 + c^2) = 16 \times \frac{81}{16} = 81$$

The answer is 81.