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Вычислим \(A=\lim \limits\_{n\to +\infty } (\sqrt{4n^2+6n -1}-\sqrt{4n^2+2n+5})\) .

Решение.

Преобразуем выражение

\(\sqrt{4n^2+6n-1}-\sqrt{4n^2+2n+5}=\dfrac{(\sqrt{4n^2+6n-1}-\sqrt{4n^2+2n+5})(\sqrt{4n^2+6n-1}+\sqrt{4n^2+2n+5})}{\sqrt{4n^2+6n-1}+\sqrt{4n^2+2n+5}}=\dfrac{(4n^2+6n-1)-(4n^2+2n+5)}{\sqrt{4n^2+6n-1}+\sqrt{4n^2+2n+5} }=\dfrac{4n-6}{\sqrt{4n^2+6n-1}+\sqrt{4n^2+2n+5}}=\dfrac{4-\dfrac{6}{n}}{\sqrt{4+\dfrac{6}{n}-\dfrac{1}{n^2}}+\sqrt{4+\dfrac{2}{n}+\dfrac{5}{n^2}}}\) .

Так как \(\lim \limits\_{n\to +\infty } \left( 4-\dfrac{6}{n}\right) =4\) , а по теореме о пределе суммы и по свойству \(\lim \limits\_{n\to +\infty } \sqrt{y\_n}=\sqrt{\lim \limits\_{n\to +\infty } y\_n}\)

\(\lim \limits\_{n\to +\infty } \left( \sqrt{4+\dfrac{6}{n}-\dfrac{1}{n^2}}+\sqrt{4+\dfrac{2}{n}+\dfrac{5}{n^2}}\right) =\sqrt{\lim \limits\_{n\to +\infty } \left( 4+\dfrac{6}{n}-\dfrac{1}{n^2}\right) }+\sqrt{\lim \limits\_{n\to +\infty } \left( 4+\dfrac{5}{n}+\dfrac{5}{n^2}\right) }=2+2=4\) ,

то по теореме о пределе частного \(A=\dfrac{4}{4}=1\) .

Ответ: \(1\) .

\(((\sqrt{4n^2+6n-1}-\sqrt{4n^2+2n+5})(\sqrt{4n^2+6n-1}+\sqrt{4n^2+2n+5})) / (\sqrt{4n^2+6n-1}+\sqrt{4n^2+2n+5})\)

[ \(((\sqrt{4n^2+6n-1}-\sqrt{4n^2+2n+5})(\sqrt{4n^2+6n-1}+\sqrt{4n^2+2n+5})) / (\sqrt{4n^2+6n-1}+\sqrt{4n^2+2n+5})\) | \(((\sqrt{4n^2+6n-1}-\sqrt{4n^2+2n+5})(\sqrt{4n^2+6n-1}+\sqrt{4n^2+2n+5})) / (\sqrt{4n^2+6n-1}+\sqrt{4n^2+2n+5})\) |123]