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  4. If displaystyle lim x arrow 0 (f(x)/x2) exists finitely and displaystyle lim x arrow 0(1+x+(f(x)/x))1 / x=e3, then ∫ f(x) log e x d x is equal to

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

Integrals

Solution:

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