Question

Sa se calculeze limita sirului:
$\lim_{n \to \infty} (\sqrt{n^{2}+\sqrt{n^{2}+1 } } -n)$

Answer 1

$\lim_{n\to \infty }\left( \tfrac{\left( \sqrt{n {}^{2} + \sqrt{n {}^{2} + 1 } } - n \right)\left(\sqrt{n {}^{2} + \sqrt{n {}^{2} + 1 } } - n\right)}{ \sqrt{n {}^{2} + \sqrt{n {}^{2} + 1 } } + n } \right) \\ \\ \\ \\ \lim _{n\to \infty }\left( \tfrac{n {}^{2} + \sqrt{n {}^{2} + 1} - n {}^{2} }{ \sqrt{n {}^{2} + \sqrt{n {}^{2} + 1 } } + n} \right) \\ \\ \\ \\ \lim _{n\to \infty }\left ( \tfrac{ \sqrt{n {}^{2} + 1 } }{ \sqrt{n {}^{2} + \sqrt{n {}^{2} + 1 } } + n } \right) \\ \\ \\ \\ \lim_{n\to \infty }\left( \tfrac{ \sqrt{n {}^{2} \left(1 + \frac{1}{n {}^{2} }\right) } }{ \sqrt{n {}^{2} \left(1 + \sqrt{ \frac{1}{n {}^{2} } + \frac{1}{n {}^{4} } } \right) + n} } \right) \\ \\ \\ \\ \lim_{n\to \infty }\left( \tfrac{n \sqrt{1 + \frac{1}{n {}^{2} } } }{n\left( \sqrt{1 + \sqrt{\frac{1}{n {}^{2} } + \frac{1}{n {}^{4} } } } + 1\right)} \right) \\ \\ \\ \\ \lim_{n\to \infty }\left( \tfrac{ \sqrt{1 + \frac{1}{n {}^{2} } } }{\sqrt{1 + \sqrt{\frac{1}{n {}^{2} } + \frac{1}{n {}^{4} } } } + 1} \right) \\ \\ \\ \\ \lim_{n\to \infty}\left( \sqrt{1 + \frac{1}{n {}^{2} } } \right) \over\lim_{n\to\infty}\left( \sqrt{1 + \sqrt{ \frac{1}{n {}^{2} } + \frac{1}{n {}^{4} } } } + 1 \right) \\ \\ \\ \\ \sqrt{\lim_{n\to \infty }\left(1 + \frac{1}{n {}^{2} } \right)}\over\lim_{n\to \infty }\left( \sqrt{1 + \sqrt{ \frac{1}{n {}^{2} } + \frac{1}{n {}^{4} } } } \right) + \lim_{n\to \infty }(1) \\ \\ \\ \\ \sqrt{\lim_{n\to \infty} (1) + \lim _{n\to \infty } \left( \frac{1}{n {}^{2} } \right)} \over \sqrt{\lim_{n\to \infty }\left(1 + \sqrt{ \frac{1}{n {}^{2} } + \frac{1}{n {}^{4} } } \right) } + 1 \\ \\ \\ \\ \sqrt{1 + 0} \over \sqrt{\lim_{n\to \infty }(1) + \lim_{n\to \infty }\left( \sqrt{ \frac{1}{n {}^{2} } + \frac{1}{n {}^{4} } } \right) } + 1 \\ \\ \\ \\ \sqrt{1 + 0} \over \sqrt{1 + \sqrt{\lim_{n\to \infty } \left( \frac{1}{n {}^{2} } + \frac{1}{n {}^{4} } \right)} } + 1 \\ \\ \\ \\ \sqrt{1 + 0} \over \sqrt{1 + \sqrt{\lim_{n\to \infty }( \frac{1}{n {}^{2} } ) + \lim_{n\to \infty }( \frac{1}{n {}^{4} } )} } + 1 \\ \\ \\ \\ = \frac{\sqrt{1 +0 }} { \sqrt{1 + \sqrt{0 + 0} } + 1 }\\ \\ \\ \\ = \frac{1}{1 + 1} \\ \\ \\ \\ \color{red} = \frac{1}{2}$