Question

If $\underset{x\to 0}{\lim }\frac{f(x)}{x^2}$ exists finitely and $\underset{x\to 0}{\lim }{\left(1+x+\frac{f(x)}{x}\right)}^{1/x}=e^3$,
then $\int f(x){\log }_{e}xdx$ is equal to

• A. $\frac{2}{3}{x}^3\left({\log }_{e}x-\frac{1}{3}\right)+c$
• B. $\frac{x^3}{3}\left({\log }_{e}x-\frac{1}{3}\right)+c$
• C. $\frac{2}{3}{x}^3\left({\log }_{e}x+1\right)+c$
• D. $\frac{2}{3}{x}^3\left({\log }_{e}x-1\right)+c$

Answer 1

Answer: (A)

We have $\underset{x\to 0}{\lim }{\left(1+x+\frac{f(x)}{x}\right)}^{1/x}=e^3$
$\Rightarrow \underset{x\to 0}{\lim }{\left(1+x\left(1+\frac{f(x)}{x^2}\right)\right)}^{1/x}=e^3$
$\Rightarrow \underset{x\to 0}{\lim }{\left[{\left(1+\frac{x^2+f(x)}{x}\right)}^{\frac{x}{x^2+f(x)}}\right]}^{\frac{x^2+f(x)}{x}\frac{1}{x}}=e^3$
$\Rightarrow \frac{x^2+f(x)}{x^2}=3$
$\Rightarrow f(x)=2x^2$
Therefore
$\int f(x){\log }_{e}xdx=2\int x^2{\log }_{e}xdx$
$=2\left[\frac{x^3}{3}{\log }_{e}x-\int \frac{x^3}{3}\cdot \frac{1}{x}dx\right]($ By Parts $)$
$=\frac{2}{3}{x}^3{\log }_{e}x-\frac{2}{9}{x}^3+c$
$=\frac{2}{3}{x}^3\left({\log }_{e}x-\frac{1}{3}\right)+c$