If x 0limax^2e^x-b _e(1+x)+cxe^-xx^2 x=1, then 16(a^2+b^2+c^2) is equal to ________.

Question

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

Answer: 81

Accepted Answer

$\underset{x	o 0}{lim}\frac{a{x}^{2}(1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}+....)-b(x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-.....)+cx(1-x+\frac{{x}^{2}}{x!}-\frac{{x}^{3}}{3!}+.....)}{{x}^{3}\, .\, \frac{\sin x}{x}}$

$=\underset{x	o \infty}{lim}\frac{(c-b)x+(\frac{b}{2}-c+a){x}^{2}+(a-\frac{b}{3}+\frac{c}{2}){x}^{3}+\ldots \ldots }{{x}^{3}}=1$

$\begin{gathered} c-b=0,\, \frac{b}{2}-c+a=0 \ a-\frac{b}{3}+\frac{c}{2}=1\, a=\frac{3}{4}\, b=c=\frac{3}{2} \ {a}^{2}+{b}^{2}+{c}^{2}=\frac{9}{16}+\frac{9}{4}+\frac{9}{4} \ 16({a}^{2}+{b}^{2}+{c}^{2})=81 \end{gathered}$

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