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Question

a)

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

$\mathop {\lim }\limits_{x \to 1}\frac{x^3-\sqrt{3x-2}}{x-1}$

c)

$\mathop {\lim }\limits_{x \to 0}\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}$

Nero · Answer 1 · 4/28/2014 9:18:16 PM

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c.

$\mathop {\lim }\limits_{x \to 0}\frac{2\sqrt{1+x}-2+2-\sqrt[3]{8-x}}{x}=\mathop {\lim }\limits_{x \to 0}(\frac{2\sqrt{1+x}-2}{x}+\frac{2-\sqrt[3]{8-x}}{x})$

$=\mathop {\lim }\limits_{x \to 0}(\frac{4x}{x(2\sqrt{1+x}+2)}+\frac{x}{x[4+2\sqrt[3]{8-x}+(\sqrt[3]{8-x})^2]})$

$=\mathop {\lim }\limits_{x \to 0}(\frac{4}{2\sqrt{1+x}+2}+\frac{1}{4+2\sqrt[3]{8-x}+(\sqrt[3]{8-x})^2})$

$=\mathop {\lim }\limits_{x \to 0}(\frac{4}{2+2}+\frac{1}{4+4+4})=\frac{13}{12}$

Trả lời 28-04-14 09:18 PM

Nero
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Nero · Answer 2 · 4/28/2014 9:08:18 PM

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b.

$\mathop {\lim }\limits_{x \to 1}\frac{x^6-3x+2}{(x-1)(x^3+\sqrt{3x-2})}=\mathop {\lim }\limits_{x \to 1}\frac{(x-1)(x^5+x^4+x^3+x^2+x-2)}{(x-1)(x^3+\sqrt{3x-2})}$

$=\mathop {\lim }\limits_{x \to 1}\frac{x^5+x^4+x^3+x^2+x-2}{x^3+\sqrt{3x-2}}=\mathop {\lim }\limits_{x \to 1}\frac{1+1+1+1+1-2}{1+1}=\frac{3}{2}$

Trả lời 28-04-14 09:08 PM

Nero
1 1

299K 319K

Nero · Answer 3 · 4/28/2014 9:00:01 PM

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a.

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

$=\mathop {\lim }\limits_{x \to 1}(\frac{2(x-1)}{(x-1)[(\sqrt[4]{2x-1})^3+(\sqrt[4]{2x-1})^2+\sqrt[4]{2x-1}+1]}+\frac{x-1}{(x-1)[(\sqrt[5]{x-2})^4-(\sqrt{x-2})^3+(\sqrt[5]{x-2})^2-\sqrt[5]{x-2}+1)})$