Tính các giới hạn:
                      
 a) $\mathop {\lim }\dfrac{6+10+14+...+\left(4n+2\right)}{n^2+4}$       b) $\mathop {\lim }\left(2n-\sqrt{4n^2+n}\right)$

 c) $\mathop {\lim }\left(\sqrt{n^2+4n}-n+2\right)$          d) $\mathop {\lim }\left(\sqrt{n^2+n+1}-\sqrt[3]{n^3+n^2}\right)$


d)
$D = \lim ( \sqrt{n^2+n+1}-n)- \lim (\sqrt[3]{n^3+n^2}-n) $
$D = \lim  \dfrac{n+1}{\sqrt{n^2+n+1}+n}- \lim\dfrac{n^2}{\sqrt[3]{(n^3+n^2)^2}+n^2+n\sqrt[3]{n^3+n^2}}  $
 $D = \lim  \dfrac{1+1/n}{\sqrt{1+1/n+1/n^2}+1}- \lim\dfrac{1}{\sqrt[3]{(1+1/n)^2}+1+\sqrt[3]{1+1/n}}  $
$D= \dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}.$

Cần trả +100vỏ sò để xem nội dung lời giải này

Anh ghi a) làm em tưởng câu a) mở ra xem mà anh treo thưởng nữa :(( –  Xusint 22-02-13 11:20 PM
câu c) em ơi –  Trần Nhật Tân 22-02-13 11:18 PM
Ủa đây đâu phải đề câu a) đâu anh –  Xusint 22-02-13 11:18 PM
Em chưa hiểu lắm ạ. –  Xusint 22-02-13 11:17 PM
b)
$B = \lim (2n - \sqrt{4n^2+n})=\lim \dfrac{4n^2-(4n^2+n)}{2n +\sqrt{4n^2+n}}=\lim \dfrac{-n}{2n +\sqrt{4n^2+n}}$
$=\lim \dfrac{-1}{2 +\sqrt{4+1/n}}=-\dfrac{1}{4}$.
uh đúng rồi em ơi. Cộng trừ nhân chia nhầm xấu hổ quá :) –  Trần Nhật Tân 23-02-13 05:30 PM
lim=-1/4 anh Tân ơi ^^ –  Melody wind 23-02-13 05:27 PM

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