Question

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

Answer 1

The limit expression is:

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

Expand each term in the numerator around $x = 0$:

For $ax^e$, use the Taylor expansion:

$ax^e = ax \left( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \right).$

For $b \log_e(1 + x)$, use the expansion $\log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$:

$b \log_e(1 + x) = b \left( x - \frac{x^2}{2} + \frac{x^3}{3} + \cdots \right).$

For $cxe^{-x}$, use the expansion $e^{-x} = 1 - x + \frac{x^2}{2!} - \cdots$:

$cxe^{-x} = cx \left( 1 - x + \frac{x^2}{2} - \cdots \right).$

Substitute these expansions into the numerator:

$\lim_{x \to 0} \frac{\left( a - b + c \right) x + \left( \frac{b}{2} - c + a \right) x^2 + \left( \frac{a}{3} - \frac{b}{2} + \frac{c}{2} \right) x^3 + \cdots}{x^3 \sin x}.$

Since $\sin x \approx x$ as $x \to 0$, we rewrite the expression as:

$\lim_{x \to 0} \frac{a - b + c + \left( \frac{b}{2} - c + a \right) x + \left( \frac{a}{3} - \frac{b}{2} + \frac{c}{2} \right) x^2 + \cdots}{x^2} = 1.$

By matching terms, we get:

$c = b = 0, \quad \frac{b}{2} - c + a = 0.$

Solving these equations, we find:

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

Calculate $a^2 + b^2 + c^2$:

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

Thus, $16(a^2 + b^2 + c^2) = 81.$