Mathematics - Limits Question with Solution | TestHub

Given: 

$\underset{x\to 0}{\lim }\frac{a{x}^2{e}^x-b{\log }_e\left(1+x\right)+cx{e}^{-x}}{x^2\sin x}=1$

$\Rightarrow \underset{x\to 0}{\lim }\frac{a{x}^2{e}^x-b{\log }_e\left(1+x\right)+cx{e}^{-x}}{x^2·x·{\displaystyle \frac{\sin x}{x}}}=1$

$\Rightarrow \underset{x\to 0}{\lim }\frac{a{x}^2\left(1+{\displaystyle \frac{x}{1!}}+{\displaystyle \frac{x^2}{2!}}+...\right)-b\left(x-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^3}{3}}+...\right)+cx\left(1-\frac{x}{1!}+\frac{x^2}{2!}+...\right)}{x^2\times x}=1$

Coefficient of $x=0$

$\Rightarrow -b+c=0 ...\left(iii\right)$

Coefficient of $x^2=0$

$\Rightarrow a+\frac{b}{2}-c=0 ...\left(ii\right)$

$\Rightarrow a-\frac{c}{2}=0$

$\Rightarrow a=\frac{c}{2}$

Coefficient of $x^3=1$

$\Rightarrow a-\frac{b}{3}+\frac{c}{2}=1 ...\left(iii\right)$

$\Rightarrow \frac{c}{2}-\frac{c}{3}+\frac{c}{2}=1$

$\Rightarrow \frac{2c}{3}=1$

$\Rightarrow c=\frac{3}{2}=b$

$\Rightarrow a=\frac{3}{4}$

$\Rightarrow 16\left(a^2+b^2+c^2\right)=16\left(\frac{9}{16}+\frac{9}{4}\times 2\right)$

$\Rightarrow 16\left(a^2+b^2+c^2\right)=81$