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Tìm các giới hạn

1> $\mathop {\lim }\limits_{x \to -\infty }$$2x+1+\sqrt{4x^2-3} $ 2> $\mathop {\lim }\limits_{x \to \frac{-\Pi }{4}}$ $\frac{sinx+cosx}{4x+\Pi }$

3> $\mathop {\lim }\limits_{x \to 2}$ $\frac{\sqrt[3]{2x-3} -\sqrt{x-1} }{x-2}$

hanhphucnhe989 · Answer 1 · 4/22/2013 3:52:44 PM

Bình chọn tăng 2 Bình chọn giảm

3> $\mathop {\lim }\limits_{x \to 2}$ $\frac{\sqrt[3]{2x-3} -\sqrt{x-1} }{x-2}=\mathop {\lim }\limits_{x \to 2}\frac{(\sqrt[3]{2x-3}-1)-(\sqrt{x-1}-1)}{x-2}$

$=\mathop {\lim }\limits_{x \to 2}\left(\frac{\sqrt[3]{2x-3}-1}{x-2}-\frac{\sqrt{x-1}-1}{x-2}\right)$

$=\mathop {\lim }\limits_{x \to 2}\left[\frac{2x-4}{(x-2)[(\sqrt[3]{2x-3})^2+\sqrt[3]{2x-3}+1}-\frac{x-2}{(x-2)(\sqrt{x-1}+1)}\right]$

$=\mathop {\lim }\limits_{x \to 2}\left[\frac{2}{(\sqrt[3]{2x-3})^2+\sqrt[3]{2x-3}+1}-\frac{1}{\sqrt{x-1}+1}\right]$

$=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$

Trả lời 22-04-13 03:52 PM

hanhphucnhe989
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nếu thấy đúng thì vote up nhé – hanhphucnhe989 23-04-13 07:32 PM

Trần Nhật Tân · Answer 2 · 4/22/2013 11:37:40 AM

Bình chọn tăng 4 Bình chọn giảm

1. $\mathop {\lim }\limits_{x \to -\infty }(2x+1+\sqrt{4x^2-3} )=\mathop {\lim }\limits_{x \to -\infty }x\left ( 2+\frac{1}{x}-\sqrt{4-\frac{3}{x^2}} \right ) =(-\infty).2=-\infty.$

Trả lời 22-04-13 11:37 AM