Friday, 15 April 2016

real analysis - Calculate limnto+infty(sqrtn2+nn)











Could someone help me through this problem?

Calculate lim


Answer



We have:



\sqrt{n^{2}+n}-n=\frac{(\sqrt{n^{2}+n}-n)(\sqrt{n^{2}+n}+n)}{\sqrt{n^{2}+n}+n}=\frac{n}{\sqrt{n^{2}+n}+n}
Therefore:



\sqrt{n^{2}+n}-n=\frac{1}{\sqrt{1+\frac{1}{n}}+1}



And since: \lim\limits_{n\to +\infty}\frac{1}{n}=0




It follows that:



\boxed{\,\,\lim\limits_{n\to +\infty}(\sqrt{n^{2}+n}-n)=\dfrac{1}{2}\,\,}


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