Em là tám lại ạ

Em là duy khôi ạ

Em là văn tam ạ

Em là mạnh Tuấn ạ

a: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x+4-12}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x-8}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x+4}{x^2+2x+4}\)

\(=\dfrac{2+4}{2^2+2\cdot2+4}=\dfrac{6}{4+4+4}=\dfrac{6}{12}=\dfrac{1}{2}\)

b: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)

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

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

\(=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{2}{\left(2-3\right)\left(2-1\right)}=-2\)

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

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

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

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

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

=0

c: \(\lim\limits_{x\rightarrow+\infty}\left[x\cdot\left(\sqrt{x^2+1}-x\right)\right]\)

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

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x}{\sqrt{x^2+1}+x}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)

e: \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{x^2+1-1}{\sqrt{x^2+1}+1}:\dfrac{x^2+16-16}{\sqrt{x^2+16}+4}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+16}+4}{\sqrt{x^2+1}+1}=\dfrac{4+4}{1+1}=\dfrac{8}{2}=4\)

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

Lời giải:

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

b) 

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

\(=+\infty\)

a: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)

\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)

b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

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

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

\(a=\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)\left(x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}\right)}{\left(x^3-3x-2\right)\left(x+\sqrt[]{x+2}\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+1\right)\left(x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}\right)}{\left(x-2\right)\left(x+1\right)^2\left(x+\sqrt[]{x+2}\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}}{\left(x+1\right)\left(x+\sqrt[]{x+2}\right)}=...\)

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

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

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

\(=...\)

\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(\sqrt[]{5+4x}-2x-3\right)+\left(2x+3-\sqrt[3]{7+6x}\right)}{x^3+x^2-x-1}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{5+4x-\left(2x+3\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(2x+3\right)^3-\left(7+6x\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4\left(x+1\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(x+1\right)^2\left(8x+20\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4}{2x+3+\sqrt[]{5+4x}}+\dfrac{8x+20}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{x-1}\)

\(=...\)

1/ \(\lim\limits_{x\rightarrow0^-}\left(\dfrac{x-2}{x^3}\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2-x}{-x^3}=\dfrac{2}{0}=+\infty\)

2/ \(\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^3-x^2\right)^{\dfrac{1}{2}}}{\left(x-1\right)^{\dfrac{1}{2}}+1-x}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^3-x^2\right)^{-\dfrac{1}{2}}.\left(3x^2-2x\right)}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}-1}=0\)

3/ \(\lim\limits_{x\rightarrow1^+}\dfrac{1-\left(x^2+x+1\right)}{x^3-1}=\dfrac{1-3}{0}=-\infty\)

4/ \(\lim\limits_{x\rightarrow-\infty}\left(-\infty-\sqrt[3]{1+\infty}\right)=-\left(\infty+\infty\right)=-\infty?\) Cái này ko chắc :v