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The limit $\displaystyle \lim_{n \rightarrow \infty} \left(\sqrt{n^{2}+n}-n\right)$ equals.

  1. $\infty$
  2. $1$
  3. $1 / 2$
  4. $0$
  5. None of the above
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3 Answers

Best answer
17 votes
17 votes

$L =\displaystyle \lim_{n \to \infty} \sqrt{n^2 + n} - n$
$\quad=\displaystyle \lim_{n\to\infty}\left(\sqrt{n^2+n}-n \right) \times \left(\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}\right)$
$\quad=\displaystyle \lim_{n\to \infty}\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}$
$\quad=\displaystyle \lim_{n\to \infty}\frac{n}{n\left(\sqrt{1+\frac{1}{n}}+1\right)}$
$\quad= \displaystyle\lim_{n\to \infty}\frac{1}{\sqrt{1+\frac{1}{n}}+1}$
$\quad=  \frac{1}{\sqrt{1 + \frac1 \infty} + 1}$
$L = \frac 1 2$

Hence, option C is the correct answer.

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6 votes
6 votes
Ans c
$\lim_{n \to \infty} (\sqrt{n^2 + n} - n)$
= $\lim_{n \to \infty} n(\sqrt{1 + \frac{1}{n}} - 1)$
= $\lim_{n \to \infty} \frac{(\sqrt{1 + \frac{1}{n}} - 1)}{\frac{1}{n}}$
= $\lim_{n \to \infty} \frac{(-n^{2})}{2\sqrt{1 + \frac{1}{n}}} *(\frac{-1}{n^{2}})$
= 1/2
–3 votes
–3 votes
=n(sqrt(1+1/n)-1)

as n tends to infinity,

=infinity*(sqrt(1+0)-1)

=0+0+0+....................infinity times

=0
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