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NT
10 tháng 11 lúc 13:28

\(...=\lim\limits_{n\rightarrow\infty}\dfrac{\sqrt{n^2\left(4+\dfrac{3}{n^2}\right)}}{n}=\)\(\lim\limits_{n\rightarrow\infty}\dfrac{\left|n\right|\sqrt{4+\dfrac{3}{n^2}}}{n}\)

\(=\left[{}\begin{matrix}\lim\limits_{n\rightarrow+\infty}\dfrac{n\sqrt{4+\dfrac{3}{n^2}}}{n}=\dfrac{2}{1}=2\\\lim\limits_{n\rightarrow-\infty}\dfrac{-n\sqrt{4+\dfrac{3}{n^2}}}{n}=-\dfrac{2}{1}=-2\end{matrix}\right.\)

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