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Oblicz granicę \(\lim_{n \to \infty} \sqrt{n^2+4n+1}-\sqrt{n^2+2n}\)
\(1\)
\[ \begin{split} &\lim_{n \to \infty} \sqrt{n^2+4n+1}-\sqrt{n^2+2n}=\\[12pt] &=\lim_{n \to \infty} \frac{n^2+4n+1-n^2-2n}{\sqrt{n^2+4n+1}+\sqrt{n^2+2n}}=\\[16pt] &=\lim_{n \to \infty} \frac{2n+1}{\sqrt{n^2+4n+1}+\sqrt{n^2+2n}}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\dfrac{2n}{n}+\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{4n}{n^2}+\dfrac{1}{n^2}}+\sqrt{\dfrac{n^2}{n^2}+\dfrac{2n}{n^2}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{2+\dfrac{1}{n}}{\sqrt{1+\dfrac{4}{n}+\dfrac{1}{n^2}}+\sqrt{1+\dfrac{2}{n}}}=\\[16pt] &=\frac{2}{\sqrt{1}+\sqrt{1}}=\frac{2}{2}=1 \end{split} \]
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