Question

Assertion :If $f\left(x\right)=\frac{1}{n}{[(n+1)(n+2)(n+3)...(n+n)]}^{\frac{1}{n}}$ then ${lim}_{n\to \infty }f\left(x\right)$ equals $\frac{4}{e}$ Reason: $\underset{n\to \infty }{lim}\frac{1}{n}f\left(\frac{r}{n}\right)={\int }_0^{1}f\left(x\right)dx$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Let $A=\underset{n\to \infty }{lim}f\left(x\right)$
$=\underset{n\to \infty }{lim}\frac{1}{n}{[(n+1)(n+2)(n+3)...(n+n)]}^{\frac{1}{n}}$
$=\underset{n\to \infty }{lim}{[\frac{n+1}{n}.\frac{n+2}{n}...\left(\frac{n+n}{n}\right)]}^{\frac{1}{n}}$
$=\underset{n\to \infty }{lim}{[(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n})]}^{\frac{1}{n}}$
$\Rightarrow$ $\log A=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\log (1+\frac{r}{n})={\int }_0^1\log (1+x)dx$
$={\left(x\log (1+x)\right)}_0^1-{\int }_0^1\frac{x}{1+x}dx$
$=\log 2-{\int }_0^1(1-\frac{1}{1+x})dx$
$=\log 2-1+\log 2=\log 4-\log e$
$\log A=\log \left(\frac{4}{e}\right)$ $\therefore$ $A=\frac{4}{e}$