Question

[Bùi Chí Minh]

a/ \(^{lim}_{x->0}\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}\)

b/\(^{lim}_{x->1}\left(\frac{1}{1-x}-\frac{1}{1-x^3}\right)\)

c/ \(^{lim}_{x->+\infty}\left(\sqrt[3]{2x-1}-\sqrt[3]{2x+1}\right)\)

d/ \(^{lim}_{x->-\infty}\left(\sqrt[3]{3x^3-1}+\sqrt{x^2+2}\right)\)

e/\(^{lim}_{x->2}\left(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}\right)\)

f/ \(^{lim}_{x->0^{+-}}\left(\frac{2x}{\sqrt{4x^2+x^3}}\right)\)

Answer 1

[Nguyễn Việt Lâm]

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.

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

Answer 2

[Bùi Chí Minh]

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