If lim _{x rightarrow 0} frac{a x^2 e^x - b l | vedclass

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(C) Given $\lim _{x \rightarrow 0} \frac{a x^2 e^x - b \log _e(1+x) + c x e^{-x}}{x^2 \sin x} = 1$. Using Taylor series expansions: $e^x = 1 + x + \fr

Vedclass · Accepted Answer

$81$

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(C) Given $\lim _{x \rightarrow 0} \frac{a x^2 e^x - b \log _e(1+x) + c x e^{-x}}{x^2 \sin x} = 1$. Using Taylor series expansions: $e^x = 1 + x + \frac{x^2}{2!} + \dots$,$\log _e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,$e^{-x} = 1 - x + \frac{x^2}{2!} - \dots$,and $\sin x \approx x$. The expression becomes $\lim _{x}$ ${\rightarrow 0} \frac{a x^2(1+x+\dots) - b(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots) + c x(1 - x + \frac{x^2}{2} - \dots)}{x^3} = 1$. Grouping terms by powers of $x$: $\lim _{x \rightarrow 0} \frac{(c-b)x + (a + \frac{b}{2} - c)x^2 + (a - \frac{b}{3} + \frac{c}{2})x^3 + \dots}{x^3} = 1$. For the limit to exist and equal $1$,coefficients of $x$ and $x^2$ must be $0$: $c - b = 0 \implies c = b$. $a + \frac{b}{2} - c = 0 \implies a + \frac{b}{2} - b = 0 \implies a = \frac{b}{2}$. Coefficient of $x^3$ must be $1$: $a - \frac{b}{3} + \frac{c}{2} = 1$. Substituting $a = \frac{b}{2}$ and $c = b$: $\frac{b}{2} - \frac{b}{3} + \frac{b}{2} = 1 \implies b - \frac{b}{3} = 1 \implies \frac{2b}{3} = 1 \implies b = \frac{3}{2}$. Thus,$c = \frac{3}{2}$ and $a = \frac{3}{4}$. Then $16(a^2 + b^2 + c^2) = 16((\frac{3}{4})^2 + (\frac{3}{2})^2 + (\frac{3}{2})^2) = 16(\frac{9}{16} + \frac{9}{4} + \frac{9}{4}) = 9 + 36 + 36 = 81$.

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

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