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This question is my doubt which has been attached by me.

This question is my doubt which has been attached by me.

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Grade:12

1 Answers

Riddhish Bhalodia
askIITians Faculty 434 Points
8 years ago
The limit is of the form 1^\infty
so we can write the limit as
L = lim_{n \rightarrow \infty}(\frac{\sqrt{n^2+n} -1}{n})^{2\sqrt{n^2+n}-1} = e^{lim_{n \rightarrow \infty}(\frac{\sqrt{n^2+n} -1}{n} - 1)({2\sqrt{n^2+n}-1)}}
so we find
L1 = {lim_{n \rightarrow \infty}(\frac{\sqrt{n^2+n} -1}{n} - 1)({2\sqrt{n^2+n}-1)}} = lim_{n \rightarrow \infty}\frac{(2n^2+3n-1) - (2n+3)\sqrt{n^2+n}}{n}

thus
L1 = lim_{n \rightarrow \infty}\frac{(2n^2+3n-1) - (2n^2+3n)\sqrt{1+1/n}}{n}
using the expansion of the under root we can easily find the limit to be
L1 = -1
and hence
L = 1/e

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