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Differential Calculus> This question is my doubt which has been ...

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

Nipurn jain , 10 Years ago

Grade 12

1 Answers

Riddhish Bhalodia

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
$L = 1/e$

Last Activity: 10 Years ago

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