toán đại số 11

Question

Tính các giới hạn :

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

3) $\mathop {\lim }\limits_{x \to 0}$$\frac{\sqrt[m]{1+ax}\sqrt[n]{1+bx}-1}{x}$ 4) $\mathop {\lim }\limits_{x \to 1}$$\frac{2-\sqrt{2x-1}\sqrt[3]{5x+3}}{x-1}$

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

score 5 · Answer 1

Bình chọn tăng 5 Bình chọn giảm

6.
$\mathop {\lim }\limits_{x \to 1}\dfrac{\sqrt{4x+5}+\sqrt{3x+1}-5}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{(\sqrt{4x+5}-3)+(\sqrt{3x+1}-2)}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{\dfrac{4x-4}{\sqrt{4x+5}+3}+\dfrac{3x-3}{\sqrt{3x+1}+2}}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\left(\dfrac{4}{\sqrt{4x+5}+3}+\dfrac{3}{\sqrt{3x+1}+2}\right)=\dfrac{17}{12}$

Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! – khangnguyenthanh 08-02-13 10:00 PM

score 5 · Answer 2

Bình chọn tăng 5 Bình chọn giảm

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

Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! – khangnguyenthanh 08-02-13 10:00 PM

score 5 · Answer 3

Bình chọn tăng 5 Bình chọn giảm

4.
$\mathop {\lim }\limits_{x \to 1}\dfrac{2-\sqrt{2x-1}\sqrt[3]{5x+3}}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{(2-2\sqrt{2x-1})+\sqrt{2x-1}(2-\sqrt[3]{5x+3})}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\dfrac{\dfrac{2(2-2x)}{1+\sqrt{2x-1}}+\dfrac{\sqrt{2x-1}(5-5x)}{4+2\sqrt[3]{5x+3}+\sqrt[3]{(5x+3)^2}}}{x-1}$
$=\mathop {\lim }\limits_{x \to 1}\left(\dfrac{-4}{1+\sqrt{2x-1}}-\dfrac{5\sqrt{2x-1}}{4+2\sqrt[3]{5x+3}+\sqrt[3]{(5x+3)^2}}\right)=\dfrac{-29}{12}$

Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! – khangnguyenthanh 08-02-13 09:59 PM

score 5 · Answer 4

Bình chọn tăng 5 Bình chọn giảm

3.
$\mathop {\lim }\limits_{x \to 0}\dfrac{\sqrt[m]{1+ax}\sqrt[n]{1+bx}-1}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\sqrt[m]{1+ax}(\sqrt[n]{1+bx}-1)+(\sqrt[m]{1+ax}-1)}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\dfrac{bx\sqrt[m]{1+ax}}{\sqrt[n]{(1+bx)^{n-1}}+\ldots+\sqrt[n]{1+bx}+1}+\dfrac{ax}{\sqrt[m]{(1+ax)^{m-1}}+\ldots+\sqrt[m]{1+ax}+1}}{x}$
$=\mathop {\lim }\limits_{x \to 0}\left(\dfrac{b\sqrt[m]{1+ax}}{\sqrt[n]{(1+bx)^{n-1}}+\ldots+\sqrt[n]{1+bx}+1}+\dfrac{a}{\sqrt[m]{(1+ax)^{m-1}}+\ldots+\sqrt[m]{1+ax}+1}\right)=\dfrac{a}{m}+\dfrac{b}{n}$

Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! – khangnguyenthanh 08-02-13 09:58 PM

score 5 · Answer 5

Bình chọn tăng 5 Bình chọn giảm

2.
$\mathop {\lim }\limits_{x \to 0}\dfrac{\sqrt{1+2x}\sqrt[3]{1+3x}\sqrt[3]{1+4x}-1}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\sqrt{1+2x}\sqrt[3]{1+3x}(\sqrt[3]{1+4x}-1)+\sqrt{1+2x}(\sqrt[3]{1+3x}-1)+(\sqrt{1+2x}-1)}{x}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\dfrac{4x\sqrt{1+2x}\sqrt[3]{1+3x}}{\sqrt[3]{(1+4x)^2}+\sqrt[3]{1+4x}+1}+\dfrac{3x\sqrt{1+2x}}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}+\dfrac{2x}{\sqrt{1+2x}+1}}{x}$
$=\mathop {\lim }\limits_{x \to 0}\left(\dfrac{4\sqrt{1+2x}\sqrt[3]{1+3x}}{\sqrt[3]{(1+4x)^2}+\sqrt[3]{1+4x}+1}+\dfrac{3\sqrt{1+2x}}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}+\dfrac{2}{\sqrt{1+2x}+1}\right)=\dfrac{10}{3}$

a có thể giải thích cho e cách làm của bài này đc k ? – mackhue59 11-02-13 12:49 AM

Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! – khangnguyenthanh 08-02-13 09:57 PM

score 5 · Answer 6

Bình chọn tăng 5 Bình chọn giảm

1.
$\mathop {\lim }\limits_{x \to 0}\dfrac{\sqrt{1-2x}-\sqrt[3]{1+3x}}{x^2}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{(\sqrt{1-2x}-1)-(\sqrt[3]{1+3x}-1)}{x^2}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\dfrac{-2x}{\sqrt{1-2x}+1}-\dfrac{3x}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}}{x^2}$
$=\mathop {\lim }\limits_{x \to 0}\dfrac{\dfrac{-2}{\sqrt{1-2x}+1}-\dfrac{3}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}}{x}$
Mà $\mathop {\lim }\limits_{x \to 0^+}\dfrac{\dfrac{-2}{\sqrt{1-2x}+1}-\dfrac{3}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}}{x}=-\infty$
$\mathop {\lim }\limits_{x \to 0^-}\dfrac{\dfrac{-2}{\sqrt{1-2x}+1}-\dfrac{3}{\sqrt[3]{(1+3x)^2}+\sqrt[3]{1+3x}+1}}{x}=+\infty$
nên giới hạn cần tìm 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 08-02-13 09:57 PM