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Question

Phương pháp: Hằng số vắng

Tính các giới hạn:
1, $\mathop {\lim }\limits_{x \to 0}$$\frac{2\sqrt{1+x} - \sqrt[3]{8-x}}{x}$

2,$\mathop {\lim }\limits_{x \to 1}$$\frac{\sqrt[4]{2x-1} + \sqrt[5]{x-2}}{x-1}$

3,$\mathop {\lim }\limits_{x \to 1}$$\frac{2\sqrt{5-x^{3}}-\sqrt[3]{x^2 +7}}{x^2 -1}$

4, $\mathop {\lim }\limits_{x \to 0}$$\frac{(x^2+2009)\sqrt[7]{1-2x}-2009}{x}$

5,$\mathop {\lim }\limits_{x \to 7}$$\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$

6, $\mathop {\lim }\limits_{x \to 1}$$\frac{\sqrt{2x-1}+x^2-3x+1}{\sqrt[3]{x-2}+x^2-x+1}$

score 5 · Answer 1

Bình chọn tăng 5 Bình chọn giảm

6.
$\mathop {\lim }\limits_{x \to 1} \dfrac{\sqrt{2x-1}+x^2-3x+1}{\sqrt[3]{x-2}+x^2-x+1}$
$=\mathop {\lim }\limits_{x \to 1} \dfrac{(\sqrt{2x-1}-1)+x^2-3x+2}{(1-\sqrt[3]{2-x})+x^2-x}$
$=\mathop {\lim }\limits_{x \to 1} \dfrac{\dfrac{2x-2}{\sqrt{2x-1}+1}+(x-1)(x-2)}{\dfrac{x-1}{1+\sqrt[3]{2-x}+\sqrt[3]{(2-x)^2}}+x(x-1)}$
$=\mathop {\lim }\limits_{x \to 1} \dfrac{\dfrac{2}{\sqrt{2x-1}+1}+x-2}{\dfrac{1}{1+\sqrt[3]{2-x}+\sqrt[3]{(2-x)^2}}+x}=0$

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score 5 · Answer 2

Bình chọn tăng 5 Bình chọn giảm

5.
$\mathop {\lim }\limits_{x \to 7}\dfrac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$
$=\mathop {\lim }\limits_{x \to 7}\dfrac{(\sqrt{x+2}-3)-(\sqrt[3]{x+20}-3)}{\sqrt[4]{x+9}-2}$
$=\mathop {\lim }\limits_{x \to 7}\dfrac{\dfrac{x-7}{\sqrt{x+2}+3}-\dfrac{x-7}{\sqrt[3]{(x+20)^2}+3\sqrt{x+20}+9}}{\dfrac{x-7}{\sqrt[4]{(x+9)^3}+2\sqrt[4]{(x+9)^2}+4\sqrt[4]{x+9}+8}}$
$=\mathop {\lim }\limits_{x \to 7}\dfrac{\dfrac{1}{\sqrt{x+2}+3}-\dfrac{1}{\sqrt[3]{(x+20)^2}+3\sqrt{x+20}+9}}{\dfrac{1}{\sqrt[4]{(x+9)^3}+2\sqrt[4]{(x+9)^2}+4\sqrt[4]{x+9}+8}}=\dfrac{112}{27}$

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score 6 · Answer 3

Bình chọn tăng 6 Bình chọn giảm

4.
$\mathop {\lim }\limits_{x \to 0}\dfrac{(x^2+2009)\sqrt[7]{1-2x}-2009}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{x^2\sqrt[7]{1-2x}+2009(\sqrt[7]{1-2x}-1)}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{x^2\sqrt[7]{1-2x}+\dfrac{-2009.2x}{\sqrt[7]{(1-2x)^6}+\sqrt[7]{(1-2x)^5}+\sqrt[7]{(1-2x)^4}+\sqrt[7]{(1-2x)^3}+\sqrt[7]{(1-2x)^2}+\sqrt[7]{1-2x}+1}}{x}$
$=\mathop {\lim }\limits_{x \to 0}\left(x\sqrt[7]{1-2x}-\dfrac{2009.2}{\sqrt[7]{(1-2x)^6}+\sqrt[7]{(1-2x)^5}+\sqrt[7]{(1-2x)^4}+\sqrt[7]{(1-2x)^3}+\sqrt[7]{(1-2x)^2}+\sqrt[7]{1-2x}+1}\right)=-574$

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score 6 · Answer 4

Bình chọn tăng 6 Bình chọn giảm

3.
$\mathop {\lim }\limits_{x \to 1^+}\dfrac{2\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}=+\infty$
$\mathop {\lim }\limits_{x \to 1^-}\dfrac{2\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}=-\infty$
$\Rightarrow \mathop {\lim }\limits_{x \to 1}\dfrac{2\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}$ không tồn tại.

Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! – khangnguyenthanh 07-02-13 09:31 AM

score 6 · Answer 5

Bình chọn tăng 6 Bình chọn giảm

2.
$\mathop {\lim }\limits_{x \to 1}\dfrac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{(\sqrt[4]{2x-1}-1)+(1-\sqrt[5]{2-x})}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{\dfrac{2x-2}{\sqrt[4]{(2x-1)^3}+\sqrt[4]{(2x-1)^2}+\sqrt[4]{2x-1}+1}+\dfrac{x-1}{\sqrt[5]{(2-x)^4}+\sqrt[5]{(2-x)^3}+\sqrt[5]{(2-x)^2}+\sqrt[5]{2-x}+1}}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\left(\dfrac{2}{\sqrt[4]{(2x-1)^3}+\sqrt[4]{(2x-1)^2}+\sqrt[4]{2x-1}+1}+\dfrac{1}{\sqrt[5]{(2-x)^4}+\sqrt[5]{(2-x)^3}+\sqrt[5]{(2-x)^2}+\sqrt[5]{2-x}+1}\right)=\dfrac{7}{10}$

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score 6 · Answer 6

Bình chọn tăng 6 Bình chọn giảm

1.
$\mathop {\lim }\limits_{x \to 0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{2(\sqrt{1+x}-1)-(\sqrt[3]{8-x}-2)}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\dfrac{2x}{\sqrt{1+x}+1}-\dfrac{-x}{\sqrt[3]{(8-x)^2}+2\sqrt{8-x}+4}}{x}$
$=\mathop {\lim }\limits_{x \to 0}\left(\dfrac{2}{\sqrt{1+x}+1}+\dfrac{1}{\sqrt[3]{(8-x)^2}+2\sqrt{8-x}+4}\right)=\dfrac{13}{12}$

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