Key Concepts and Formulas

• Taylor Series Expansions: For small values of $x$, we can use the Taylor series expansions of common functions around $x=0$:
  • $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
  • $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$
  • $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
• Limit Properties: If a limit of a rational function (or a ratio of series) exists, the coefficients of the terms that would otherwise lead to an indeterminate form (like $\frac{0}{0}$) must satisfy certain conditions. Specifically, for the limit to be finite, the numerator's terms of the lowest power must cancel out until a non-zero term remains in the numerator that matches the lowest power in the denominator.

Step-by-Step Solution

Step 1: Expand the functions in the numerator using Taylor series. To evaluate the limit, we will substitute the Taylor series expansions for $e^x$, $\log_e(1+x)$, and $\sin x$ around $x=0$. We need to expand them to a sufficient order to determine the coefficients.

• $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + O(x^4)$
• $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$
• $\sin x = x - \frac{x^3}{3!} + O(x^5)$

Step 2: Substitute the Taylor series into the given limit expression. Now, we substitute these expansions into the numerator and the denominator.

The numerator is $a x^2 e^x - b \log_e(1+x) + c x e^{-x}$. Let's expand $e^{-x}$: $e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + O(x^4)$.

Numerator: $a x^2 \left(1 + x + \frac{x^2}{2} + O(x^3)\right) - b \left(x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)\right) + c x \left(1 - x + \frac{x^2}{2} + O(x^3)\right)$

Expanding this: $ax^2 + ax^3 + \frac{a}{2}x^4 + O(x^5) - bx + \frac{b}{2}x^2 - \frac{b}{3}x^3 + O(x^4) + cx - cx^2 + \frac{c}{2}x^3 + O(x^4)$

Collecting terms by powers of $x$: $(-b+c)x + \left(a+\frac{b}{2}-c\right)x^2 + \left(a-\frac{b}{3}+\frac{c}{2}\right)x^3 + O(x^4)$

The denominator is $x^2 \sin x$. Substituting the series for $\sin x$: $x^2 \left(x - \frac{x^3}{6} + O(x^5)\right) = x^3 - \frac{x^5}{6} + O(x^7)$

So the limit becomes: $\lim _{x \rightarrow 0} \frac{(-b+c)x + \left(a+\frac{b}{2}-c\right)x^2 + \left(a-\frac{b}{3}+\frac{c}{2}\right)x^3 + O(x^4)}{x^3 + O(x^5)}$

Step 3: Analyze the limit and set up equations for the coefficients. For the limit to be a finite value (specifically, 1), the terms in the numerator with powers of $x$ less than the lowest power in the denominator (which is $x^3$) must be zero. This is because if there were a non-zero term of $x$ or $x^2$ in the numerator, the limit would go to 0 or $\pm \infty$.

• Coefficient of $x$: $-b+c = 0$
• Coefficient of $x^2$: $a+\frac{b}{2}-c = 0$
• Coefficient of $x^3$: For the limit to be 1, the coefficient of $x^3$ in the numerator must be equal to the coefficient of $x^3$ in the denominator (which is 1). So, $a-\frac{b}{3}+\frac{c}{2} = 1$.

Step 4: Solve the system of linear equations for $a$, $b$, and $c$. We have the following system of equations:

1. $c - b = 0 \implies c = b$
2. $a + \frac{b}{2} - c = 0$
3. $a - \frac{b}{3} + \frac{c}{2} = 1$

Substitute $c=b$ into equation (2): $a + \frac{b}{2} - b = 0$ $a - \frac{b}{2} = 0 \implies a = \frac{b}{2}$

Now substitute $c=b$ and $a=\frac{b}{2}$ into equation (3): $\frac{b}{2} - \frac{b}{3} + \frac{b}{2} = 1$ $b - \frac{b}{3} = 1$ $\frac{3b-b}{3} = 1$ $\frac{2b}{3} = 1 \implies b = \frac{3}{2}$

Now we can find $a$ and $c$: $c = b = \frac{3}{2}$ $a = \frac{b}{2} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$

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

Step 5: Calculate the value of $16(a^2+b^2+c^2)$. Now we substitute the values of $a$, $b$, and $c$ into the expression:

$a^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$ $b^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$ $c^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$

$a^2+b^2+c^2 = \frac{9}{16} + \frac{9}{4} + \frac{9}{4}$ To add these fractions, we find a common denominator, which is 16: $a^2+b^2+c^2 = \frac{9}{16} + \frac{9 \times 4}{4 \times 4} + \frac{9 \times 4}{4 \times 4}$ $a^2+b^2+c^2 = \frac{9}{16} + \frac{36}{16} + \frac{36}{16}$ $a^2+b^2+c^2 = \frac{9 + 36 + 36}{16} = \frac{81}{16}$

Finally, calculate $16(a^2+b^2+c^2)$: $16\left(a^2+b^2+c^2\right) = 16 \times \frac{81}{16} = 81$.

Common Mistakes & Tips

• Incorrect Taylor Series Expansion: Ensure you use the correct Taylor series for each function and expand to a sufficient order (at least up to $x^3$ in this case, as the denominator is $x^3$).
• Algebraic Errors: Solving the system of linear equations can be prone to arithmetic mistakes. Double-check your calculations.
• Denominator Power: Be mindful of the lowest power of $x$ in the denominator. For the limit to be finite, all terms in the numerator with powers lower than this must cancel out.

Summary

The problem involves evaluating a limit of a rational function where the numerator and denominator both tend to zero as $x \to 0$. We used Taylor series expansions for $e^x$, $\log_e(1+x)$, and $\sin x$ around $x=0$ to rewrite the expression. For the limit to be a finite non-zero value, the coefficients of the terms with powers of $x$ less than the lowest power in the denominator ($x^3$) must be zero. This yielded a system of three linear equations in $a$, $b$, and $c$. Solving this system gave us the values of $a$, $b$, and $c$. Finally, we substituted these values into the expression $16(a^2+b^2+c^2)$ to find the required result.

The final answer is \boxed{81}.