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NV
6 tháng 6 2020

\(\lim\limits_{x\rightarrow1}\frac{\sqrt{5x-1}-2+2-\sqrt[3]{10x-2}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{5\left(x-1\right)}{\sqrt{5x-1}+2}-\frac{10\left(x-1\right)}{4+2\sqrt[3]{10x-2}+\sqrt[3]{\left(10x-2\right)^2}}}{\sqrt{2}\left(x-1\right)}\)

\(=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{2}}\left(\frac{5}{\sqrt{5x-1}+2}-\frac{10}{4+2\sqrt[3]{10x-2}+\sqrt[3]{\left(10x-2\right)^2}}\right)=\frac{5}{12\sqrt{2}}=\frac{5}{3.2^2\sqrt{2}}\)

\(\Rightarrow\left\{{}\begin{matrix}a=5\\b=2\end{matrix}\right.\) \(\Rightarrow a+b=7\)

NV
15 tháng 3 2020

\(a=\lim\limits_{x\rightarrow a}\frac{\left(\sqrt{x}-\sqrt{a}\right)\left(x+\sqrt{ax}+a\right)}{\sqrt{x}-\sqrt{a}}=\lim\limits_{x\rightarrow a}\left(x+\sqrt{ax}+a\right)=3a\)

\(b=\lim\limits_{x\rightarrow1}\frac{x^{\frac{1}{n}}-1}{x^{\frac{1}{m}}-1}=\lim\limits_{x\rightarrow1}\frac{\frac{1}{n}x^{\frac{1-n}{n}}}{\frac{1}{m}x^{\frac{1-m}{m}}}=\frac{\frac{1}{n}}{\frac{1}{m}}=\frac{m}{n}\)

Ta có:

\(\lim\limits_{x\rightarrow1}\frac{1-\sqrt[n]{x}}{1-x}=\lim\limits_{x\rightarrow1}\frac{1-x^{\frac{1}{n}}}{1-x}=\lim\limits_{x\rightarrow1}\frac{-\frac{1}{n}x^{\frac{1-n}{n}}}{-1}=\frac{1}{n}\)

\(\Rightarrow c=\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)}{1-x}.\frac{\left(1-\sqrt[3]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[4]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)}=\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}=\frac{1}{120}\)

\(d=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}+1}=\frac{1}{2}\)

NV
15 tháng 3 2020

\(e=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{x}{\sqrt{1+x}+1}+\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\frac{1}{\sqrt{1+x}+1}+\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}\right)=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

\(f=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{3+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\)

\(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{3+\sqrt{x+7}}}{x-1}=\frac{8}{27}-\frac{1}{6}=\frac{7}{54}\)

\(g=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-1+1-\sqrt{2x-1}}{\left(x-1\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{3\left(x-1\right)}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2\left(x-1\right)}{1+\sqrt{2x-1}}}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\frac{\frac{3}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2}{1+\sqrt{2x-1}}}{x^2+x+1}=0\)

\(h=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}=\frac{\sqrt[3]{10}-\sqrt[3]{4}}{2}\)

NV
15 tháng 3 2020

\(a=\lim\limits_{x\rightarrow-\infty}\left(\frac{-x^2}{\sqrt[3]{\left(x^3-x^2\right)^2}+x\sqrt[3]{x^3-x^2}+x^2}\right)=\lim\limits_{x\rightarrow-\infty}\left(\frac{-1}{\sqrt[3]{\left(1-\frac{1}{x}\right)^3}+\sqrt[3]{1-\frac{1}{x}}+1}\right)=-\frac{1}{3}\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{5x^2-8x}{\sqrt[3]{\left(x^3+5x^2\right)^2}+\sqrt[3]{\left(x^3+5x^2\right)\left(x^3+8x\right)}+\sqrt[3]{\left(x^3+8x\right)^2}}\)

\(=\lim\limits_{x\rightarrow+\infty}\frac{5-\frac{8}{x}}{\sqrt[3]{\left(1+\frac{5}{x}\right)^2}+\sqrt[3]{\left(1+\frac{5}{x}\right)\left(1+\frac{8}{x^2}\right)}+\sqrt[3]{\left(1+\frac{8}{x^2}\right)^2}}=\frac{5}{3}\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{1}{\sqrt[3]{\left(x^3+1\right)^2}+x\sqrt[3]{x^3+1}+x^2}=\frac{1}{+\infty}=0\)

Bài 2:

\(a=\lim\limits_{x\rightarrow1^-}\left(\frac{1-x}{\left(x-1\right)\left(x+1\right)}\right)=\lim\limits_{x\rightarrow1^-}\frac{-1}{x+1}=-\frac{1}{2}\)

\(b=\lim\limits_{x\rightarrow1^+}\left(\frac{x^2+x+1-3}{\left(1-x\right)\left(x^2+x+1\right)}\right)=\lim\limits_{x\rightarrow1^+}\frac{\left(x-1\right)\left(x+2\right)}{\left(1-x\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1^+}\frac{-x-2}{x^2+x+1}=-1\)

\(c=\lim\limits_{x\rightarrow2^+}\left(\frac{1}{\left(x-1\right)\left(x-2\right)}-\frac{1}{\left(x-2\right)\left(x-3\right)}\right)=\lim\limits_{x\rightarrow2^+}\frac{-2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)

Do \(x\rightarrow2^+\Rightarrow x>2\Rightarrow x-2>0\Rightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\rightarrow0^-\)

\(\Rightarrow\lim\limits_{x\rightarrow2^+}\frac{-2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=+\infty\)

NV
22 tháng 1

\(a+\dfrac{x+1}{\sqrt{x^2-x+1}}-\dfrac{3x+3}{\sqrt{x}}=0\) có nghiệm \(x=1\)

\(\Rightarrow a+\dfrac{2}{\sqrt{1}}-\dfrac{6}{\sqrt{1}}=0\Rightarrow a=4\)

\(4+\dfrac{x+1}{\sqrt{x^2-x+1}}-\dfrac{3x+3}{\sqrt{x}}=3\left(2-\dfrac{x+1}{\sqrt{x}}\right)+\left(\dfrac{x+1}{\sqrt{x^2-x+1}}-2\right)\)

\(=-3\left(\dfrac{\left(x-1\right)^2}{\sqrt{x}\left(x+1+2\sqrt{x}\right)}\right)+\dfrac{-3\left(x-1\right)^2}{\sqrt{x^2-x+1}\left(x+1-2\sqrt{x^2-x+1}\right)}\)

Rút gọn với \(\left(x-1\right)^2\) bên ngoài rồi thay dố là được

5 tháng 3 2019

Nếu m,n là số tự nhiên :v

\(x^n-1=\left(x-1\right)\left(x^{n-1}+x^{n-2}+...+x+1\right)\)\(x^m-1=\left(x-1\right)\left(x^{m-1}+x^{m-2}+...+x+1\right)\)

\(\lim\limits_{x\rightarrow1}\left(\frac{x^m-1}{x^n-1}\right)=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^{m-1}+...+x+1\right)}{\left(x-1\right)\left(x^{n-1}+...+x+1\right)}\)

\(=\frac{x^{m-1}+...+x+1}{x^{n-1}+....+x+1}=\frac{m}{n}\)

5 tháng 3 2019

Cảm ơn cậu nhiều ạ. Nếu có thể giúp tớ câu b được không. Xin cảm ơn.

NV
15 tháng 2 2020

a/ Do \(x\rightarrow-3^+\) nên \(x>-3\Rightarrow x+3>0\Rightarrow\left|x+3\right|=x+3\)

\(\Rightarrow\lim\limits_{x\rightarrow-3^+}\frac{3x+9}{\left|x+3\right|}=\lim\limits_{x\rightarrow-3^+}\frac{3\left(x+3\right)}{x+3}=3\)

b/ \(=\lim\limits_{x\rightarrow0^+}\frac{\sqrt{x}\left(1-3\sqrt{x}\right)}{\sqrt{x}\left(4\sqrt{x}-2\right)}=\lim\limits_{x\rightarrow0^+}\frac{1-3\sqrt{x}}{4\sqrt{x}-2}=-\frac{1}{2}\)

Ở câu này \(x\rightarrow0^+\) có nghĩa \(x>0\), nó chỉ để căn thức xác định, ngoài ra ko có gì đặc biệt hết

c/ Tương tự câu c, cũng chỉ để căn thức xác định \(\left(x< 1\right)\)

\(\lim\limits_{x\rightarrow1^-}\frac{\sqrt{1-x}}{\left(1-x\right)\left(x+4\right)}=\lim\limits_{x\rightarrow1^-}\frac{1}{\sqrt{1-x}\left(x+4\right)}=+\infty\)

d/ Chắc bạn ghi nhầm đề, đây ko phải giới hạn dạng vô định (vì tử khác 0, mẫu bằng 0):

\(x\rightarrow\sqrt{2}^-\Rightarrow x< \sqrt{2}\Rightarrow x^4-4< 0\)

\(\Rightarrow\lim\limits_{x\rightarrow\sqrt{2}^-}\frac{\left|x-2\right|}{x^4-4}=-\infty\)

AH
Akai Haruma
Giáo viên
25 tháng 1 2020

Lời giải:
a)

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

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

b)

\(\lim\limits_{x\to1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(\sqrt[3]{x}-1\right)^2}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(x-1\right)^2}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2\left(x-1\right)^2}\)

\(=\lim\limits_{x\to1}\frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2}=\frac{1}{\left(1+1+1\right)^2}=\frac{1}{9}\)

c)

\(\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}=\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{(x-1)(x^2+2x+2)}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x-1)(x^2+2x+2)}\)

\(=\lim_{x\to 1}\frac{1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x^2+2x+2)}=\frac{1}{(1+1)(1+1)(1+2.1+2)}=\frac{1}{20}\)

d)

\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)

\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)

\(=\frac{-7}{12}\)

AH
Akai Haruma
Giáo viên
7 tháng 1 2020

Lời giải:
a)

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

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

b)

\(\lim\limits_{x\to1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(\sqrt[3]{x}-1\right)^2}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(x-1\right)^2}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2\left(x-1\right)^2}\)

\(=\lim\limits_{x\to1}\frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2}=\frac{1}{\left(1+1+1\right)^2}=\frac{1}{9}\)

c)

\(\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}=\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{(x-1)(x^2+2x+2)}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x-1)(x^2+2x+2)}\)

\(=\lim_{x\to 1}\frac{1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x^2+2x+2)}=\frac{1}{(1+1)(1+1)(1+2.1+2)}=\frac{1}{20}\)

d)

\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)

\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)

\(=\frac{-7}{12}\)

8 tháng 11 2023

\(4\sqrt{2}x\) ạ

1, \(\lim\limits_{x\rightarrow1}\frac{2x^2-3x+1}{x^3-x^2-x+1}\) 2, \(\lim\limits_{x\rightarrow2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}\) 3, \(\lim\limits_{x\rightarrow0}\frac{1-\sqrt[3]{x-1}}{x}\) 4, \(\lim\limits_{x\rightarrow-\infty}\frac{x^2-5x+1}{x^2-2}\) 5, \(\lim\limits_{x\rightarrow+\infty}\frac{2x^2-4}{x^3+3x^2-9}\) 6, \(\lim\limits_{x\rightarrow2^-}\frac{2x-1}{x-2}\) 7, \(\lim\limits_{x\rightarrow3^+}\frac{8+x-x^2}{x-3}\) 8,...
Đọc tiếp

1, \(\lim\limits_{x\rightarrow1}\frac{2x^2-3x+1}{x^3-x^2-x+1}\)

2, \(\lim\limits_{x\rightarrow2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}\)

3, \(\lim\limits_{x\rightarrow0}\frac{1-\sqrt[3]{x-1}}{x}\)

4, \(\lim\limits_{x\rightarrow-\infty}\frac{x^2-5x+1}{x^2-2}\)

5, \(\lim\limits_{x\rightarrow+\infty}\frac{2x^2-4}{x^3+3x^2-9}\)

6, \(\lim\limits_{x\rightarrow2^-}\frac{2x-1}{x-2}\)

7, \(\lim\limits_{x\rightarrow3^+}\frac{8+x-x^2}{x-3}\)

8, \(\lim\limits_{x\rightarrow-\infty}\left(8+4x-x^3\right)\)

9, \(\lim\limits_{x\rightarrow-1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}\)

10, \(\lim\limits_{x\rightarrow-\infty}\frac{\left(2x^2+1\right)^2\left(5x+3\right)}{\left(2x^3-1\right)\left(x+1\right)^2}\)

11, \(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{x^2+2x}}{x+3}\)

12, \(\lim\limits_{x\rightarrow1}\frac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}\)

13, \(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{x+1}+\sqrt{x+4}-3}{x}\)

14, \(\lim\limits_{x\rightarrow0}\frac{\left(x^2+2020\right)\sqrt{1+3x}-2020}{x}\)

15, \(\lim\limits_{x\rightarrow+\infty}\left(2x-\sqrt{4x^2-3}\right)\)

16, \(\lim\limits_{x\rightarrow a}\frac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)

17, \(\lim\limits_{x\rightarrow1}\frac{x^n-nx+n-1}{\left(x-1\right)^2}\)

18, \(f\left(x\right)=\left\{{}\begin{matrix}\frac{x^2-2x}{8-x^3}\\\frac{x^4-16}{x-2}\end{matrix}\right.\) khi x>2,khi x<2 tại x=2

9
AH
Akai Haruma
Giáo viên
12 tháng 3 2020

Bài 2:

\(\lim\limits_{x\to 2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}=\lim\limits_{x\to 2}\frac{x^2-x-2}{(x+\sqrt{x+2}).\frac{4x+1-9}{\sqrt{4x+1}+3}}=\lim\limits_{x\to 2}\frac{(x-2)(x+1)(\sqrt{4x+1}+3)}{(x+\sqrt{x+2}).4(x-2)}=\lim\limits_{x\to 2}\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}=\frac{9}{8}\)

Bài 3:

\(\lim\limits_{x\to 0-}\frac{1-\sqrt[3]{x-1}}{x}=-\infty \)

\(\lim\limits_{x\to 0+}\frac{1-\sqrt[3]{x-1}}{x}=+\infty \)

Bài 4:

\(\lim\limits_{x\to -\infty}\frac{x^2-5x+1}{x^2-2}=\lim\limits_{x\to -\infty}\frac{1-\frac{5}{x}+\frac{1}{x^2}}{1-\frac{2}{x^2}}=1\)

Bài 5:

\(\lim\limits_{x\to +\infty}\frac{2x^2-4}{x^3+3x^2-9}=\lim\limits_{x\to +\infty}\frac{\frac{2}{x}-\frac{4}{x^3}}{1+\frac{3}{x}-\frac{9}{x^3}}=0\)

AH
Akai Haruma
Giáo viên
12 tháng 3 2020

Bài 6:

\(\lim\limits_{x\to 2- }\frac{2x-1}{x-2}=\lim\limits_{x\to 2-}\frac{2(x-2)+3}{x-2}=\lim\limits_{x\to 2-}\left(2+\frac{3}{x-2}\right)=-\infty \)

Bài 7:

\(\lim\limits _{x\to 3+ }\frac{8+x-x^2}{x-3}=\lim\limits _{x\to 3+}\frac{1}{x-3}.\lim\limits _{x\to 3+}(8+x-x^2)=2(+\infty)=+\infty \)

Bài 8:

\(\lim\limits _{x\to -\infty}(8+4x-x^3)=\lim\limits _{x\to -\infty}(-x^3)=+\infty \)

Bài 9:

\(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\)

\(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)

NV
15 tháng 3 2020

Bài 1:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)

\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)

NV
15 tháng 3 2020

Bài 2:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)

\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)