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

Correct Answer: 81

Given:

\[ \lim_{x \to 0} \frac{a x^2 \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \right) - b \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \right) + c x \left( 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots \right)}{x^3 \cdot \frac{\sin x}{x}} \]

\[ = \lim_{x \to 0} \frac{(c - b)x + \left( \frac{b}{2} - c + a \right)x^2 + \left( a - \frac{b}{3} + \frac{c}{2} \right)x^3 + \ldots}{x^3} = 1 \]

Step 1: Equating coefficients

\[ c - b = 0, \quad \frac{b}{2} - c + a = 0 \]

\[ a - \frac{b}{3} + \frac{c}{2} = 1 \]

Step 2: Solving the equations

From \( c - b = 0 \Rightarrow b = c \)

Substituting into \( \frac{b}{2} - c + a = 0 \): \[ \frac{b}{2} - b + a = 0 \Rightarrow a = \frac{b}{2} \]

Substituting into \( a - \frac{b}{3} + \frac{c}{2} = 1 \): \[ \frac{b}{2} - \frac{b}{3} + \frac{b}{2} = 1 \] \[ \Rightarrow \frac{4b}{3} = 1 \Rightarrow b = \frac{3}{2} \]

\[ \therefore a = \frac{3}{4}, \quad b = \frac{3}{2}, \quad c = \frac{3}{2} \]

Step 3: Finding \( a^2 + b^2 + c^2 \)

\[ a^2 + b^2 + c^2 = \left(\frac{9}{16}\right) + \left(\frac{9}{4}\right) + \left(\frac{9}{4}\right) = \frac{9}{16} + \frac{18}{4} = \frac{81}{16} \]

Step 4:

\[ 16(a^2 + b^2 + c^2) = 81 \]

Final Answer:

\[ \boxed{81} \]

Answer 2

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, \[ 16(a^2 + b^2 + c^2) = 81. \]