Question

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

• A. $a=3,b=12,c=9$
• B. $a=1,b=2,c=4$
• C. $a=2,b=10,c=84$
• D. $a=3,b=-12,c=-9$

Answer 1

The correct option is A $a=3,b=12,c=9$

$\underset{x\to 0}{lim}\frac{ax{e}^x-b\log (1+x)+cx{e}^{-x}}{x^2\sin x}$ which is equal to $\frac{0}{0}$. Hence applying L'Hospitals rule

$=\underset{x\to 0}{lim}\frac{a[e^x+x{e}^x]-\frac{b}{(1+x)}+c[e^{-x}-x{e}^{-x}]}{2x\sin x+x^2\cos x}$

$=\frac{a-b+c}{0}$ Hence for the limit to exist $a-b+c=0$ $\Rightarrow$ (1)

Applying L'Hospitals rule again

$\underset{x\to 0}{lim}\frac{a[e^x+e^x+x{e}^x]+\frac{b}{(1+x{)}^2}+c(-e^{-x})(1-x)-c{e}^{-x}}{2\sin x+2x\cos x+2x\cos x-x^2\sin x}$

$=\frac{2a+b-2c}{0}$ Hence for the limit to exist $2a+b-2c=0$ $\Rightarrow$ (2)

Applying L'Hospitals rule again

$\underset{x\to 0}{lim}\frac{a{e}^x[x+2]+a{e}^x-\frac{2b}{(1+x{)}^3}+c{e}^{-x}(1-x)+c{e}^{-x}+c{e}^{-x}}{2\cos x+4\cos x-4x\sin x-2x\sin x-x^2\cos x}$

$=\frac{3a-2b+3c}{6}=2$ [From question]

$\therefore 3a-2b+3c=12$ $\Rightarrow$ (3)

$a-b+c=0$ $\to \left(1\right)\Rightarrow a=b-c$.

$2a+b-2c=0$ $\to \left(2\right)\Rightarrow 2b-2c+b-2c=0\Rightarrow 3b=4c\Rightarrow b=\frac{4c}{3}$

$3a-2b+3c=12$ $\to \left(3\right)\Rightarrow 3[b-c]-2b+3c=12\Rightarrow b=12$

$b=\frac{4c}{3}\Rightarrow c=12\times \frac{3}{4}=9$

$a=b-c=12-9=3$

$\therefore a=3,b=12,c=9$ [A]