\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x^4}{x^4}+\dfrac{7}{x^4}}{\dfrac{x^4}{x^4}+\dfrac{1}{x^4}}=1\)

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

b/ x tien toi duong vo cung hay am vo cung ban?

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

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

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

\(=\dfrac{\sqrt[4]{\left(2+1\right)^3}+\sqrt[4]{\left(2+1\right)^2}+\sqrt[4]{2+1}+1}{2.\left(\sqrt[3]{\left(1+1\right)^2}+\sqrt[3]{1+1}+1\right)}=....\)

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

De bai la nhu nay ha ban?

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x}-1}{x-1}?\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(3x+3-9\right)\left(x+\sqrt{x+2}\right)}{\left(x^2-x-2\right)\left(\sqrt{3x+3}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{3\left(x-2\right)\left(x+\sqrt{x+2}\right)}{\left(x-2\right)\left(x+1\right)\left(\sqrt{3x+3}+3\right)}=\dfrac{3\left(2+\sqrt{2+2}\right)}{\left(2+1\right)\left(\sqrt{3.2+3}+3\right)}=\dfrac{2}{3}\)

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

b/ \(=\lim\limits_{x\rightarrow2}\dfrac{4x+1-9}{\left(x-2\right)\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{4}{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}=\dfrac{4}{\left(2+2\right)\left(\sqrt{4.2+1}+3\right)}=\dfrac{1}{6}\)

c/ \(=\lim\limits_{x\rightarrow-2}\dfrac{2x+5-1}{\left(x-2\right)\left(x+2\right)\left(\sqrt{2x+5}+1\right)}=\lim\limits_{x\rightarrow-2}\dfrac{2}{\left(x-2\right)\left(\sqrt{2x+5}+1\right)}=\dfrac{2}{\left(-2-2\right)\left(\sqrt[2]{2.\left(-2\right)+5}+1\right)}=\dfrac{2}{\left(-4\right).2}=-\dfrac{1}{4}\)

\(f\left(-2\right)=-2m+1\)

\(\lim\limits_{x\rightarrow-2^+}f\left(x\right)=\lim\limits_{x\rightarrow-2^+}\dfrac{x^2-3x+2}{x^3+8}=\lim\limits_{x\rightarrow-2^+}\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\lim\limits_{x\rightarrow-2^+}\dfrac{x-1}{x^2-2x+4}=\dfrac{-2-1}{4-2.\left(-2\right)+4}=-\dfrac{1}{4}\)

\(f\left(-2\right)\ne\lim\limits_{x\rightarrow-2^-}f\left(x\right)\Leftrightarrow-2m+1\ne-\dfrac{1}{4}\Leftrightarrow m\ne\dfrac{5}{8}\)

\(f\left(0\right)=2.0+m+1=m+1\)

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt[3]{x+1}-1}{x}=\lim\limits_{x\rightarrow0^+}\dfrac{x+1-1}{x(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1)}=\dfrac{1}{1+1+1}=\dfrac{1}{3}\)\(f\left(0\right)=\lim\limits_{x\rightarrow0^+}f\left(x\right)\Leftrightarrow m+1=\dfrac{1}{3}\Rightarrow m=-\dfrac{2}{3}\)