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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)
\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)
\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)
\(=\frac{-7}{12}\)
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)
\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)
\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)
\(=\frac{-7}{12}\)
Xet \(m\ne-3\)
\(=\lim\limits_{x\rightarrow-\infty}x\left(\sqrt[3]{1}+\sqrt{4}+m\right)=x\left(3+m\right)\)
\(=\left[{}\begin{matrix}-\infty\left(m>-3\right)\\+\infty\left(m< -3\right)\end{matrix}\right.\)
Xet \(m=-3\)
\(=\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+2x^2+1}-x-2x-\sqrt{4x^2+2x+3}\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^3+2x^2+1-x^3}{\sqrt[3]{\left(x^3+2x^2+1\right)^2}+x\sqrt[3]{x^3+2x^2+1}+x^2}-\lim\limits_{x\rightarrow-\infty}\dfrac{4x^2-4x^2-2x-3}{2x-\sqrt{4x^2+2x+3}}\)
\(=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\)
Bạn bị nhầm số rồi. Xét $m>1; m< 1; m=1$ mới đúng chứ
Nếu
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+x+1}-2\sqrt{x^2-x+1}\right)=\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}-2\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}\right)\)
\(=+\infty.\left(1-2\right)=-\infty\)
Nếu:
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2+x+1}-2\sqrt{x^2-x+1}\right)=\lim\limits_{x\rightarrow-\infty}x\left(-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+2\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}\right)\)
\(=-\infty.\left(-1+2\right)=-\infty\)
\(a=\lim\limits_{x\rightarrow0}\frac{x^2}{x\left(\sqrt{1+x^2}+1\right)}=\lim\limits_{x\rightarrow0}\frac{x}{\sqrt{1+x^2}+1}=\frac{0}{2}=0\)
\(b=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}=\lim\limits_{x\rightarrow1}\frac{\frac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{\left(x-1\right)\left(x+1\right)}{2+\sqrt{5-x^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\frac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{x+1}{2+\sqrt{5-x^2}}\right)=\frac{1}{12}+\frac{1}{2}=\frac{7}{12}\)
\(c=\lim\limits_{x\rightarrow0}\frac{2x}{x\left(\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}\right)}=\lim\limits_{x\rightarrow0}\frac{2}{\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}}=\frac{2}{3}\)
\(d=\frac{\sqrt[3]{6}}{0}=+\infty\)
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}-x}+\lim\limits_{x\rightarrow-\infty}\dfrac{3x^3-1-x^3}{\sqrt[3]{\left(3x^3-1\right)^2}+x\sqrt[3]{3x^3-1}+x^2}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}+\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{1}{x^2}}{\dfrac{\sqrt[3]{\left(3x^3-1\right)^2}}{x^2}+\dfrac{x\sqrt[3]{3x^3-1}}{x^2}+\dfrac{x^2}{x^2}}=0\)
b/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x-x^2}{\sqrt{x^2+x}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^3-x^3+x^2}{x^2+x\sqrt[3]{x^3-x^2}+\sqrt[3]{\left(x^3-x^2\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}+\dfrac{x}{x}}+\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x^2}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{x\sqrt[3]{x^3-x^2}}{x^2}+\dfrac{\sqrt[3]{\left(x^3-x^2\right)^2}}{x^2}}\)
\(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x-1-2x-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{4x^2-1}+\sqrt[3]{\left(2x+1\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2}{x^{\dfrac{2}{3}}}}{\dfrac{\sqrt[3]{\left(2x-1\right)^2}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{4x^2-1}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{\left(2x+1\right)^2}}{x^{\dfrac{2}{3}}}}=0\)
Check lai ho minh nhe :v
\(a=\frac{0-1}{0-1}=1\)
\(b=\lim\limits_{x\rightarrow0}\frac{\frac{x^2}{\sqrt[3]{\left(1+x^2\right)^2}+\sqrt[3]{1+x^2}+1}}{x^2}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt[3]{\left(1+x^2\right)^2}+\sqrt[3]{1+x^2}+1}=\frac{1}{3}\)
\(c=\lim\limits_{x\rightarrow2}\frac{\sqrt{x+2}-2+\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{\frac{x-2}{\sqrt{x+2}+2}+\frac{x-2}{\sqrt{x+7}+3}}{x-2}=\lim\limits_{x\rightarrow2}\left(\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{x+7}+3}\right)\)
\(=\frac{1}{\sqrt{4}+2}+\frac{1}{\sqrt{9}+3}=\frac{5}{12}\)
Vậy nó ko phải dạng vô định, cứ thay số trực tiếp
\(=\frac{2}{0}=+\infty\)
Nếu là mũ 3 thì nó là dạng 0/0 rút gọn được. Nên chắc là đề ghi nhầm đấy
