JEE Mains · Maths · STD 12 - 7.2 definite integral
If \(\mathrm{U}_{\mathrm{n}}=\left(1+\frac{1}{\mathrm{n}^{2}}\right)\left(1+\frac{2^{2}}{\mathrm{n}^{2}}\right)^{2} \ldots\left(1+\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}}\right)^{\mathrm{n}}\), then \(\lim _{n \rightarrow \infty}\left(U_{n}\right)^{\frac{-4}{n^{2}}}\) is equal to :
- A \(\frac{\mathrm{e}^{2}}{16}\)
- B \(\frac{4}{\mathrm{e}}\)
- C \(\frac{16}{\mathrm{e}^{2}}\)
- D \(\frac{4}{\mathrm{e}^{2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{e}^{2}}{16}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{U}_{\mathrm{n}}=\prod_{\mathrm{r}=1}^{\mathrm{n}}\left(1+\frac{\mathrm{r}^{2}}{\mathrm{n}^{2}}\right)^{\mathrm{r}}\) \(\mathrm{L}=\lim _{\mathrm{n} \rightarrow \infty}\left(\mathrm{U}_{\mathrm{n}}\right)^{-4 / \mathrm{n}^{2}}\)…
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