Ofcourse I can see that $\lim_{n \to \infty} \sqrt{n+1}-\sqrt{n} = 0$ just by looking aat it, but how can I prove it in the right way?
How to show that $\lim_{n \to \infty} \sqrt{n+1}-\sqrt{n} = 0$?
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$\begingroup$
limits
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0See also: https://math.stackexchange.com/questions/545704/show-that-sqrtn1-sqrtn-to0 https://math.stackexchange.com/questions/1447360/limit-problem-sqrtx1-sqrtx-as-x-approaches-infinity https://math.stackexchange.com/questions/1582826/prove-that-the-limit-of-sqrtn1-sqrtn-is-zero – 2017-04-18
2 Answers
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Hint: Use $a-b = \frac{a^2-b^2}{a+b}.$
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0Why would anyone downvote this? Just for the laughs? – 2017-01-03
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hint: Use $0 < \sqrt{n+1} - \sqrt{n} < \dfrac{1}{\sqrt{n}}$