lim(can(4n^2-2n) -2n) lim1/(can n^2+2)-(can n^2 +4) 15/11/2021 Bởi Elliana lim(can(4n^2-2n) -2n) lim1/(can n^2+2)-(can n^2 +4)
1) $\lim(\sqrt{4n^2 – 2n} – 2n)$ $=\lim\dfrac{(\sqrt{4n^2 – 2n} – 2n)(\sqrt{4n^2 – 2n} +2n)}{\sqrt{4n^2 – 2n} +2n}$ $=\lim\dfrac{-2n}{\sqrt{4n^2 – 2n} + 2n}$ $=\lim\dfrac{-2}{\sqrt{4 -\dfrac2n} +2}$ $=\dfrac{-2}{\sqrt{4-2.0} +2}$ $=-\dfrac12$ 2) $\lim\dfrac{1}{\sqrt{n^2 +2} -\sqrt{n^2 +4}}$ $=\lim\dfrac{\sqrt{n^2 +2} +\sqrt{n^2 +4}}{(\sqrt{n^2 +2} -\sqrt{n^2 +4})(\sqrt{n^2 +2} +\sqrt{n^2 +4})}$ $=-\dfrac12\lim(\sqrt{n^2 +2} +\sqrt{n^2 +4})$ $=-\dfrac12\lim\left[n\left(\sqrt{1+\dfrac{2}{n^2}} + \sqrt{1+\dfrac{4}{n^2}}\right)\right]$ $=-\dfrac12\cdot+\infty(\sqrt{1+2.0} + \sqrt{1 + 4.0})$ $= -\infty$ Bình luận
1) $\lim(\sqrt{4n^2 – 2n} – 2n)$
$=\lim\dfrac{(\sqrt{4n^2 – 2n} – 2n)(\sqrt{4n^2 – 2n} +2n)}{\sqrt{4n^2 – 2n} +2n}$
$=\lim\dfrac{-2n}{\sqrt{4n^2 – 2n} + 2n}$
$=\lim\dfrac{-2}{\sqrt{4 -\dfrac2n} +2}$
$=\dfrac{-2}{\sqrt{4-2.0} +2}$
$=-\dfrac12$
2) $\lim\dfrac{1}{\sqrt{n^2 +2} -\sqrt{n^2 +4}}$
$=\lim\dfrac{\sqrt{n^2 +2} +\sqrt{n^2 +4}}{(\sqrt{n^2 +2} -\sqrt{n^2 +4})(\sqrt{n^2 +2} +\sqrt{n^2 +4})}$
$=-\dfrac12\lim(\sqrt{n^2 +2} +\sqrt{n^2 +4})$
$=-\dfrac12\lim\left[n\left(\sqrt{1+\dfrac{2}{n^2}} + \sqrt{1+\dfrac{4}{n^2}}\right)\right]$
$=-\dfrac12\cdot+\infty(\sqrt{1+2.0} + \sqrt{1 + 4.0})$
$= -\infty$