Question

The value of $\underset{x\to 0}{lim}\frac{1}{x^3}{\int }_0^x\frac{tln(1+t)}{t^4+4}dt$ is

• A. $0$
• B. $\frac{1}{12}$
• C. $\frac{1}{24}$
• D. $\frac{1}{64}$

Answer 1

The correct option is B $\frac{1}{12}$

$\underset{x\to 0}{lim}\frac{1}{x^3}{\int }_0^x\frac{tln(1+t)}{t^4+4}dt$
By applying L'Hospital rule, we get
$\therefore \underset{x\to 0}{lim}\frac{1}{3x^2}\frac{xln(1+x)}{x^4+4}$

$\therefore \underset{x\to 0}{lim}\frac{1}{3x}\frac{ln(1+x)}{(x^4+4)}$
$\because {lim}_{x\to 0}\frac{log(1+x)}{x}=1$
So,$\underset{x\to 0}{lim}\frac{1}{3(x^4+4)}=\frac{1}{12}$