a. \(lim_{x\rightarrow3}\dfrac{x^3-27}{3x^2-5x-2}=\dfrac{3^3-27}{3.3^2-5.3-2}=\dfrac{0}{10}=0\)

b. \(lim_{x\rightarrow2}\dfrac{\sqrt{x+2}-2}{4x^2-3x-2}=\dfrac{\sqrt{2+2}-2}{4.2^2-3.2-2}=\dfrac{0}{8}=0\)

c. \(lim_{x\rightarrow1}\dfrac{1-x^2}{x^2-5x+4}=lim_{x\rightarrow1}\dfrac{\left(1-x\right)\left(x+1\right)}{\left(x-1\right)\left(x-4\right)}=lim_{x\rightarrow1}\dfrac{-\left(x+1\right)}{x-4}=\dfrac{-\left(1+1\right)}{1-4}=\dfrac{2}{3}\)

d. Câu này mình chịu, nhìn đề hơi lạ so với bình thường hehe

ai giup voi

Bạn tự hiểu là giới hạn tiến đến đâu nhé, làm biếng gõ đủ công thức

a. \(\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\frac{\frac{x}{\sqrt{1+x}+1}-\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}=\frac{1}{\sqrt{1+x}+1}-\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

b.

\(\frac{1-x^3-1+x}{\left(1-x\right)^2\left(1+x+x^2\right)}=\frac{x\left(1-x\right)\left(1+x\right)}{\left(1-x\right)^2\left(1+x+x^2\right)}=\frac{x\left(1+x\right)}{\left(1-x\right)\left(1+x+x^2\right)}=\frac{2}{0}=\infty\)

c.

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

d.

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

e.

\(=\frac{2x^2-8x+8}{\left(x-1\right)\left(x-2\right)\left(x-2\right)\left(x-3\right)}=\frac{2\left(x-2\right)^2}{\left(x-1\right)\left(x-3\right)\left(x-2\right)^2}=\frac{2}{\left(x-1\right)\left(x-3\right)}=\frac{2}{-1}=-2\)

f.

\(=\frac{2x}{x\sqrt{4+x}}=\frac{2}{\sqrt{4+x}}=1\)

cậu giúp mình bài mình mới đăng đc ko ạ

Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.

\(\mathop {\lim }\limits_{x \to  + \infty } \frac{{2{\rm{x}} - 1}}{x} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {2 - \frac{1}{x}} \right)}}{x} = \mathop {\lim }\limits_{x \to  + \infty } \left( {2 - \frac{1}{x}} \right) = 2 - 0 = 2\)

Chọn A.

\(a=\lim\limits_{x\rightarrow3}\frac{\left(x-3\right)\left(2x+3\right)}{\left(x-3\right)\left(x^3+3x^2+9x\right)}=\lim\limits_{x\rightarrow3}\frac{2x+3}{x^3+3x^2+9x}=\frac{2.3+3}{3^3+2.3^2+9.3}=...\)

\(b=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x^4+x^2+2x^3+2x+2\right)}=\frac{1+1}{1+1+2+2+2}=...\)

\(c=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)^2\left(4x^3+3x^2+2x+1\right)}{\left(x-1\right)^2\left(x^2+x+2\right)}=\frac{4+3+2+1}{1+1+2}=...\)

\(d=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{1+1+1+1+1}{1+1+1}=...\)

\(Lim_{x\rightarrow3}\frac{x^4-27x}{2x^2-3x-9}=Lim_{x\rightarrow3}\frac{x\left(x^3-3^3\right)}{\left(x-3\right)\left(2x+3\right)}\)

\(=Lim_{x\rightarrow3}\frac{x\left(x-3\right)\left(x^2+3x+9\right)}{\left(x-3\right)\left(2x+3\right)}=Lim_{x\rightarrow3}\frac{x\left(x^2+3x+9\right)}{2x+3}\)

\(=\frac{3\left(3^2+3.3+9\right)}{3.2+3}=\frac{3\left(9+9+9\right)}{9}=9\)

Vậy \(Lim_{x\rightarrow3}\frac{x^4-27x}{2x^2-3x-9}=9\)

`a)lim_{x->+oo}[x+1]/[x^2+x+1]`

`=lim_{x->+oo}[1/x+1/[x^2]]/[1+1/x+1/[x^2]]`

`=0`

`b)lim_{x->+oo}[3x+1]/[3x^2-x+5]`

`=lim_{x->+oo}[3/x+1/[x^2]]/[3-1/x+5/[x^2]]`

`=0`

`c)lim_{x->-oo}[3x+5]/[\sqrt{x^2+x}]`

`=lim_{x->-oo}[3+5/x]/[-\sqrt{1+1/x}]`

`=-3`

`d)lim_{x->+oo}[-5x+1]/[\sqrt{3x^2+1}]`

`=lim_{x->+oo}[-5+1/x]/[\sqrt{3+1/[x^2]}]`

`=-5/3`

\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\sqrt[]{2x+1}-1\right)+2-\sqrt[3]{x^2+x+8}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2.2x}{\sqrt[]{2x+1}+1}-\dfrac{x\left(x+1\right)}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{2x+1}+1}-\dfrac{x+1}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}\right)\)

\(=\dfrac{23}{12}\)

`a)lim_{x->+oo}[5x^2+x^3+5]/[4x^3+1]`       `ĐK: 4x^3+1 ne 0`

`=lim_{x->+oo}[5/x+1+5/[x^3]]/[4+1/[x^3]]`

`=1/4`

`b)lim_{x->-oo}[2x^2-x+1]/[x^3+x-2x^2]`      `ĐK: x ne 0;x ne 1`

`=lim_{x->-oo}[2/x-1/[x^2]+1/[x^3]]/[1+1/[x^2]-2/x]`

`=0`

Câu `c` giống `b`.