If lim x→0 ax^2e^x−bloge(1+x)+cxe^−x / x^2sinx = 1, then 16(a^2 + b^2 + c^2) is equal to _______.

Question

If \(\lim\limits _{x \rightarrow 0} \frac{\mathrm{ax}^{2} \mathrm{e}^{\mathrm{x}}-\mathrm{b} \log _{\mathrm{e}}(1+\mathrm{x})+\mathrm c \mathrm{xe}^{-\mathrm{x}}}{\mathrm{x}^{2} \sin \mathrm{x}}=1\), then \(16\left(\mathrm{a^{2}+b^{2}+c^{2}}\right)\) is equal to _____.

Answer 1

Correct answer: 81

\(\lim\limits _{x \rightarrow 0} \frac{ ax^{2}\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+....\right)-b\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}- ....\right) +c x\left(1-x+\frac{x^{2}}{x !}-\frac{x^{3}}{3 !}+....\right)}{x^{3} \cdot \frac{\sin x}{x}}\)

\( =\lim\limits_{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^{2}+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^{3}+....}{x^{3}}=1\)

\( \mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0\)

\( a-\frac{b}{3}+\frac{c}{2}=1 \quad a=\frac{3}{4} \quad b=c=\frac{3}{2}\)

\(a^{2}+b^{2}+c^{2}=\frac{9}{16}+\frac{9}{4}+\frac{9}{4}\)

\(16\left(a^{2}+b^{2}+c^{2}\right)=81\)