lim _{n rightarrow infty}left[frac{n}{(n+1) s | vedclass

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(A) The given limit is $S = \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n}{(n+k) \sqrt{k(2n+k)}}$.We can rewrite the general term as:$S = \lim _{n

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$\frac{1}{(1+x) \sqrt{2x+x^2}}$

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(A) The given limit is $S = \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n}{(n+k) \sqrt{k(2n+k)}}$.We can rewrite the general term as:$S = \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n}{(n+k) \sqrt{k \cdot n \left(2+\frac{k}{n}\right)}}$$S = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{n^2}{(n+k) \sqrt{\frac{k}{n} \cdot n \left(2+\frac{k}{n}\right)}}$$S = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\left(1+\frac{k}{n}\right) \sqrt{\frac{k}{n} \left(2+\frac{k}{n}\right)}}$Using the definition of the definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n g\left(\frac{k}{n}\right) = \int_0^1 g(x) dx$,where $x = \frac{k}{n}$.Thus,$S = \int_0^1 \frac{1}{(1+x) \sqrt{x(2+x)}} dx = \int_0^1 \frac{1}{(1+x) \sqrt{2x+x^2}} dx$.Comparing this with $\int_0^1 f(x) dx$,we get $f(x) = \frac{1}{(1+x) \sqrt{2x+x^2}}$.

$\lim _{n \rightarrow \infty}\left[\frac{n}{(n+1) \sqrt{2n+1}}+\frac{n}{(n+2) \sqrt{2(2n+2)}}+\frac{n}{(n+3) \sqrt{3(2n+3)}}+\ldots n \text{ terms}\right]=\int_0^1 f(x) d x$,then $f(x)=$

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