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Lời giải:
a)
\(\lim\limits_{x\to +\infty}\frac{\sqrt[3]{x^3+2x^2-4x+1}}{\sqrt{2x^2+x-8}}=\lim\limits_{x\to +\infty}\frac{\sqrt[3]{1+\frac{2}{x}-\frac{4}{x^2}+\frac{1}{x^3}}}{\sqrt{2+\frac{1}{x}-\frac{8}{x^2}}}\)
\(=\frac{1}{\sqrt{2}}\)
b)
\(\lim\limits_{x\to -\infty}\frac{\sqrt{x^2-2x+4}-x}{3x-1}=\lim\limits_{x\to -\infty}\frac{\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}+1}{-3+\frac{1}{x}}=\frac{-1}{3}\)
`a)lim_{x->+oo} (2x-\sqrt{x^2+4x-3})` `ĐK: x < -2-\sqrt{7};x > -2+\sqrt{7}`
`=lim_{x->+oo} [x(2-\sqrt{1+4/x -3/[x^2]}]`
`=+oo`
`b)lim_{x->+oo} (\sqrt{4x^2-3x+1}-2x)`
`=lim_{x->+oo} [4x^2-3x+1-4x^2]/[\sqrt{4x^2-3x+1}+2x]`
`=lim_{x->+oo} [-3x+1]/[\sqrt{4x^2-3x+1}+2x]`
`=lim_{x->+oo} [-3+1/x]/[\sqrt{4-3/x+1/[x^2]}+2]`
`=-3/4`
`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/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x}{x}+\dfrac{3}{x}}{\dfrac{3x}{x}-\dfrac{1}{x}}=\dfrac{1}{3}\)
b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{4}{x^2}}-\dfrac{x}{x}}{\dfrac{3x}{x}-\dfrac{1}{x}}=-\dfrac{2}{3}\)
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/ \(=\lim\limits_{x\rightarrow4^-}\dfrac{5-2x}{4-x}=-\infty\)
b/ \(=\lim\limits_{x\rightarrow+\infty}x^3\left(-1\right)=-\infty\)
1/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-x^2-x-x}{x+\sqrt{x^2+x+1}}=\dfrac{-2}{1-1}=-\infty\)
2/ tien toi +- vo cung?
3/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{8x^3+2x-8x^3}{\sqrt[3]{\left(8x^3+2x\right)^2}+2x.\sqrt[3]{8x^3+2x}+4x^2}=\dfrac{\dfrac{2x}{x^2}}{\dfrac{4x^2}{x^2}+\dfrac{4x^2}{x^2}+\dfrac{4x^2}{x^2}}=0\)
4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{16x^4+3x+1-16x^4}{\sqrt[4]{\left(16x^4+3x+1\right)^3}+2x.\sqrt[4]{\left(16x^4+3x+1\right)^2}+4x^2.\sqrt[4]{16x^4+3x+1}+8x^3}+\lim\limits_{x\rightarrow+\infty}\dfrac{4x^2-4x^2-2}{2x+\sqrt{4x^2+2}}=\dfrac{\dfrac{3x}{x^3}}{8+8+8+8}-\dfrac{\dfrac{2}{x}}{2+2}=0\)
5/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x-x^2}{\sqrt{x^2-x}+x}=\dfrac{\dfrac{1}{x}}{1+1}-\dfrac{\dfrac{x}{x}}{1+1}=-\dfrac{1}{2}\)