a/ \(\lim\limits_{x\to 1} f(x)=\frac{x^{2}-5x + 6}{x-2} \)

\(<=>\lim\limits_{x\to 1} f(x)=\dfrac{(x-3)(x-2)}{x-2} \)

<=>\(\lim\limits_{x\to 1} f(x)=x-3 \)

\(<=>\lim\limits_{x\to 1} f(x)=-2\)

`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`

a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\sqrt{\dfrac{4x^2}{x^2}-\dfrac{2}{x^2}}-x\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{1}{x^3}}}{-x\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-x}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4}-1}{-1-1}=\dfrac{3}{2}\)

b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2x}{x}+\dfrac{3}{x}}{-\sqrt{\dfrac{2x^2}{x^2}-\dfrac{3}{x^2}}}=\dfrac{2}{-\sqrt{2}}=-\sqrt{2}\)

c/ \(\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}{\dfrac{3}{x^2}-\dfrac{x^2}{x^2}}=\dfrac{2}{-1}=-2\)

`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`.

1/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{2x}{x}-\sqrt{\dfrac{3x^2}{x^2}+\dfrac{2}{x^2}}}{\dfrac{5x}{x}+\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}}=\dfrac{2-\sqrt{3}}{5+1}=\dfrac{2-\sqrt{3}}{6}\)

2/ \(=\lim\limits_{x\rightarrow+\infty}\sqrt{\dfrac{\dfrac{x^2}{x^4}+\dfrac{1}{x^4}}{\dfrac{2x^4}{x^4}+\dfrac{x^2}{x^4}-\dfrac{3}{x^4}}}=0\)

3/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt[3]{\dfrac{x^6}{x^6}+\dfrac{x^4}{x^6}+\dfrac{1}{x^6}}}{\sqrt{\dfrac{x^4}{x^4}+\dfrac{x^3}{x^4}+\dfrac{1}{x^4}}}=-1\)

Da nan roi mang meo lam mat het bai -.-

1/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{\dfrac{3x^3}{x^3}+\dfrac{1}{x^3}}+\sqrt{\dfrac{2x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}{-\sqrt[4]{\dfrac{4x^4}{x^4}+\dfrac{2}{x^4}}}=\dfrac{-\sqrt[3]{3}-\sqrt{2}}{\sqrt[4]{4}}\)

2/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{8x^7}{\left(-2x^7\right)}=-\dfrac{8}{2^7}\)

3/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4x^2-3x+4-4x^2\right)\left(\sqrt{x^2+x+1}+x\right)}{\left(x^2+x+1-x^2\right)\left(\sqrt{4x^2-3x+4}+2x\right)}=\dfrac{-3.2}{2}=-3\)

a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x}{x}+\dfrac{3}{x}}{\dfrac{3x}{x}-\dfrac{1}{x}}=\dfrac{1}{3}\)

b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{4}{x^2}}-\dfrac{x}{x}}{\dfrac{3x}{x}-\dfrac{1}{x}}=-\dfrac{2}{3}\)