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 \(\_\_\_\_\_\).

Updated On: Nov 19, 2024
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Correct Answer: 81

Solution and Explanation

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

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