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(D) The given limit is of the form $1^{\infty}$.Let $L = \lim _{n \rightarrow \infty}\left(1+\frac{H_n}{n^{2}}\right)^{n}$,where $H_n = 1+\frac{1}{2}+

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(D) The given limit is of the form $1^{\infty}$.Let $L = \lim _{n \rightarrow \infty}\left(1+\frac{H_n}{n^{2}}\right)^{n}$,where $H_n = 1+\frac{1}{2}+\ldots+\frac{1}{n}$.Using the formula $\lim _{n \rightarrow \infty}(1+f(n))^{g(n)} = e^{\lim _{n \rightarrow \infty} f(n)g(n)}$,we get:$L = \exp \left(\lim _{n \rightarrow \infty} \frac{H_n}{n^2} \cdot n\right) = \exp \left(\lim _{n \rightarrow \infty} \frac{H_n}{n}\right)$.We know that $H_n \approx \ln(n) + \gamma$ (where $\gamma$ is the Euler-Mascheroni constant).Thus,$\lim _{n \rightarrow \infty} \frac{\ln(n) + \gamma}{n} = 0$.Therefore,$L = e^0 = 1$.

$\lim _{n \rightarrow \infty}\left(1+\frac{1+\frac{1}{2}+\ldots+\frac{1}{n}}{n^{2}}\right)^{n} = \dots$

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