Question

$L=\underset{x\to 0}{lim}\frac{\sin x+a{e}^x+b{e}^{-x}+c\ln (1+x)}{x^3}=$

Equation $a{x}^2+bx+c=0$ has

• A. real and equal roots
• B. complex roots
• C. unequal positive real roots
• D. unequal roots

Answer 1

The correct option is D unequal roots

$L=\underset{x\to 0}{lim}\frac{sinx+a{e}^x+b{e}^{-x}+cl(1+x)}{x^3}$
$=\underset{x\to 0}{lim}\frac{(x-\frac{x^3}{3!}+....)+a(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+....)+b(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+....)+c(x-\frac{x^2}{2}+\frac{x^3}{3}+....)}{x^3}$
$=\underset{x\to 0}{lim}\frac{(a+b)+(1+a-b+c)x+(\frac{a}{2}+\frac{b}{2}-\frac{c}{2})x^2+(-\frac{1}{3!}+\frac{a}{3!}-\frac{b}{3!}+\frac{c}{3})x^3+....}{x^3}$
or $a+b=0,1+a-b+c=0,\frac{a}{2}+\frac{b}{2}-\frac{c}{2}=0$
and $L=-\frac{1}{3!}+\frac{a}{3!}-\frac{a}{3!}+\frac{c}{3}$
Solving the first three equations, we get $c=0,a=-1/2,b=1/2$
Then, $L=-1/3.$
The given equation $a{x}^2+bx+c=0$ reduces to $x^2-x=0$ or $x=0,1.$
Hence, option 'D' is correct.