Question

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

The value of $L$ is

• A. $1/2$
• B. $-1/3$
• C. $-1/6$
• D. $3$

Answer 1

The correct option is B $-1/3$

$L=\underset{x\to 0}{lim}\frac{sinx+a{e}^x+b{e}^{-x}+c\ln (1+x)}{x^3}$
$=\underset{x\to 0}{lim}\frac{(x-\frac{x^3}{3!}+....)+a(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+....)}{x^3}$ $\frac{+b(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+....)+c(x-\frac{x^2}{2}+\frac{x^3}{3}+....)}{x^3}$
$=\underset{x\to 0}{lim}\frac{(a+b)+(1+a-b+c)x+(\frac{a}{2}+\frac{b}{2}-\frac{c}{2})x^2}{x^3}\frac{+(-\frac{1}{3!}+\frac{a}{3!}-\frac{b}{3!}+\frac{c}{3})x^3+....}{x^3}$
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, $L=-1/3.$
Hence, option 'B' is correct.