If limx→0ax2.ex−bloge(1+x)+cxe−xx2sinx=1, then 16(a2+b2+c2) is equal to

limx→0ax2(1+x+x22!+....)−b(x−x22+x33+....)+cx(1−x+x22!−x33!.....)x3.sinxx=limx→0(c−b)x+(b2−c+a)x2(a−b3+c2)x3........x3=c−b=0,b2−c+a=0,a−b3+c2=1a=34,b=c=32

If limx→0ax2.ex−bloge(1+x)+cxe−xx2sinx=1, then 16(a2+b2+c2) is equal to

limx→0ax2(1+x+x22!+....)−b(x−x22+x33+....)+cx(1−x+x22!−x33!.....)x3.sinxx=limx→0(c−b)x+(b2−c+a)x2(a−b3+c2)x3........x3=c−b=0,b2−c+a=0,a−b3+c2=1a=34,b=c=32

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

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Detailed Solution

$\begin{array}{l} \lim_{x \to 0} \frac{a x^{2} (1 + x + \frac{x^{2}}{2!} + ....) - b (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} + ....) + c x (1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} .....)}{x^{3} . \frac{\sin x}{x}} \\ = \lim_{x \to 0} \frac{(c - b) x + (\frac{b}{2} - c + a) x^{2} (a - \frac{b}{3} + \frac{c}{2}) x^{3} ........}{x^{3}} \\ = c - b = 0, \frac{b}{2} - c + a = 0, a - \frac{b}{3} + \frac{c}{2} = 1 \\ a = \frac{3}{4}, b = c = \frac{3}{2} \end{array}$