If lim _{n rightarrow infty} frac{1}{n} log l | vedclass

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(C) Let $l = \lim _{n \rightarrow \infty} \frac{1}{n} \log \left(\frac{(2 n)!}{n^n n!}\right)$.We can write $\frac{(2n)!}{n! n^n} = \frac{(n+1)(n+2)\d

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$\log x$

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(C) Let $l = \lim _{n \rightarrow \infty} \frac{1}{n} \log \left(\frac{(2 n)!}{n^n n!}\right)$.We can write $\frac{(2n)!}{n! n^n} = \frac{(n+1)(n+2)\dots(n+n)}{n^n} = \prod_{r=1}^n \left(\frac{n+r}{n}\right) = \prod_{r=1}^n \left(1 + \frac{r}{n}\right)$.Taking the logarithm, we get $\log \left(\prod_{r=1}^n \left(1 + \frac{r}{n}\right)\right) = \sum_{r=1}^n \log \left(1 + \frac{r}{n}\right)$.Thus, $l = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1 + \frac{r}{n}\right)$.Using the definition of a definite integral as the limit of a sum, $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n g\left(\frac{r}{n}\right) = \int_0^1 g(x) dx$.Here, $g(x) = \log(1+x)$, so $l = \int_0^1 \log(1+x) dx$.Let $1+x = t$, then $dx = dt$. When $x=0, t=1$ and when $x=1, t=2$.So, $l = \int_1^2 \log t dt = \int_1^2 \log x dx$.Comparing this with $\int_1^2 f(x) dx$, we get $f(x) = \log x$.

If $\lim _{n \rightarrow \infty} \frac{1}{n} \log \left(\frac{(2 n)!}{n^n \cdot n!}\right)=\int_1^2 f(x) d x$, then $f(x)=$

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