toán đại số 11

Question

Tính các giới hạn:

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

3) $\mathop {\lim }\limits_{x \to 1}$$\frac{\sqrt{2x+2}-\sqrt[3]{7x+1}}{x-1}$ 4) $\mathop {\lim }\limits_{x \to -\infty }$ $\frac{\sqrt{x^2+2x}+3x}{\sqrt{4x^2+1}-x+2}$

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

7) $\mathop {\lim }\limits_{x \to +\infty }$ $\frac{\sqrt{x^2+4x+5}+2x+1}{\sqrt{3x^2-2x+7}+x}$ 8) $\mathop {\lim }\limits_{x \to -\infty }$$\frac{\sqrt{x^2-7x+12}}{3\left| {x} \right|-17}$

score 4 · Answer 1

Bình chọn tăng 4 Bình chọn giảm

8) $\mathop {\lim }\limits_{x \to -\infty }$$\dfrac{\sqrt{x^2-7x+12}}{3\left| {x} \right|-17}=\mathop {\lim }\limits_{x \to -\infty }\dfrac{\sqrt{x^2-7x+12}}{-3x-17}$, do $x \to -\infty\Rightarrow x < 0.$
$=\mathop {\lim }\limits_{x \to -\infty }$$\dfrac{\dfrac{\sqrt{x^2-7x+12}}{x}}{\dfrac{-3x-17}{x}}=\mathop {\lim }\limits_{x \to -\infty }\dfrac{-\sqrt{1+\dfrac{12}{x^2}-\dfrac{7}{x}}}{-\dfrac{17}{x}-3}=\dfrac{1}{3}$

score 4 · Answer 2

Bình chọn tăng 4 Bình chọn giảm

7) $\mathop {\lim }\limits_{x \to +\infty }\dfrac{\sqrt{x^2+4x+5}+2x+1}{\sqrt{3x^2-2x+7}+x}=\mathop {\lim }\limits_{x \to -\infty }\dfrac{\dfrac{\sqrt{x^2+4x+5}+2x+1}{x}}{\dfrac{\sqrt{3x^2-2x+7}+x}{x}}=\mathop {\lim }\limits_{x \to -\infty }\dfrac{-\sqrt{1+\dfrac{4}{x}+\dfrac{5}{x^2}}+2+\dfrac{1}{x}}{-\sqrt{3-\dfrac{2}{x}+\dfrac{7}{x^2}}+1}=\dfrac{1}{1-\sqrt 3}$

score 4 · Answer 3

Bình chọn tăng 4 Bình chọn giảm

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

score 4 · Answer 4

Bình chọn tăng 4 Bình chọn giảm

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

score 4 · Answer 5

Bình chọn tăng 4 Bình chọn giảm

4) $\mathop {\lim }\limits_{x \to -\infty }$ $\dfrac{\sqrt{x^2+2x}+3x}{\sqrt{4x^2+1}-x+2}$$=\mathop {\lim }\limits_{x \to -\infty }\dfrac{\dfrac{\sqrt{x^2+2x}+3x}{x}}{\dfrac{\sqrt{4x^2+1}-x+2}{x}}=\mathop {\lim }\limits_{x \to -\infty }\dfrac{-\sqrt{1+\dfrac{2}{x}}+3}{-\sqrt{4+\dfrac{1}{x^2}}-1+\dfrac{2}{x}}=-\dfrac{2}{3}$

score 4 · Answer 6

Bình chọn tăng 4 Bình chọn giảm

3) $\mathop {\lim }\limits_{x \to 1}$$\dfrac{\sqrt{2x+2}-\sqrt[3]{7x+1}}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{\sqrt{2x+2}-2}{x-1}-\mathop {\lim }\limits_{x \to 1}\dfrac{\sqrt[3]{7x+1}-2}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{2}{\sqrt{2x+2}+2}-\mathop {\lim }\limits_{x \to 1}\dfrac{7}{\sqrt[3]{(7x+1)^2}+2\sqrt[3]{7x+1}+4}$
$=\dfrac{2}{4}-\dfrac{7}{12}=-\dfrac{1}{12}$

score 4 · Answer 7

Bình chọn tăng 4 Bình chọn giảm

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

score 5 · Answer 8

Bình chọn tăng 5 Bình chọn giảm

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

Hỏi	10-02-13 11:43 PM
Lượt xem	3401
Hoạt động	11-02-13 11:40 AM

toán đại số 11

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toán đại số 11

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