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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