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\(\lim\limits_{x\rightarrow+\infty}\left(4x^2-3x+1\right)=\lim\limits_{x\rightarrow+\infty}x^2\left(4-\dfrac{3}{x}+\dfrac{1}{x^2}\right)\)
Do \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow+\infty}x^2=+\infty\\\lim\limits_{x\rightarrow+\infty}\left(4-\dfrac{3}{x}+\dfrac{1}{x^2}\right)=4>0\end{matrix}\right.\)
\(\Rightarrow\lim\limits_{x\rightarrow+\infty}x^2\left(4-\dfrac{3}{x}+\dfrac{1}{x^2}\right)=+\infty\)
1 ) \(lim_{x\rightarrow+\infty}\dfrac{3x^2+5}{x^3-x+2}=lim_{x\rightarrow+\infty}\dfrac{\dfrac{3}{x}+\dfrac{5}{x^3}}{1-\dfrac{1}{x^2}+\dfrac{2}{x^3}}=0\)
2 ) \(lim_{x\rightarrow-\infty}\dfrac{2x^2\left(3x^2-5\right)^3\left(1-x\right)^5}{3x^{14}+x^2-1}\) \(=lim_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}\left(3-\dfrac{5}{x^2}\right)^3\left(\dfrac{1}{x}-1\right)^5}{3+\dfrac{1}{x^{12}}-\dfrac{1}{x^{14}}}=0\)
3 ) \(lim_{x\rightarrow+\infty}\dfrac{3x-\sqrt{2x^2+5}}{x^2-4}=lim_{x\rightarrow+\infty}\dfrac{\left(7x^2-5\right)}{\left(3x+\sqrt{2x^2+5}\right)\left(x^2-4\right)}\)
\(=lim_{x\rightarrow+\infty}\dfrac{\dfrac{7}{x}-\dfrac{5}{x^3}}{\left(3+\sqrt{2+\dfrac{5}{x^2}}\right)\left(1-\dfrac{4}{x^2}\right)}=0\)
`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`
*** Mình nhớ là đã nhắc nhở bạn về việc sử dụng hộp công thức toán để viết đề dễ hiểu hơn. Lần nữa thì mình xin phép xóa bài nhé. Bạn sử dụng bộ gõ công thức toán ở biểu tượng $\sum$
Lời giải:
\(\lim\limits_{x\to +\infty}(\sqrt[3]{x^3+5x}-\sqrt{x^2-3x+6})=\lim\limits_{x\to +\infty}[(\sqrt[3]{x^3+5x}-x)-(\sqrt{x^2-3x+6}-x)]\)
\(=\lim\limits_{x\to +\infty}\left[\frac{5x}{\sqrt[3]{(x^3+5x)^2}+x\sqrt[3]{x^3+5x}+x^2}-\frac{-3x+6}{\sqrt{x^2-3x+6}+x}\right]\)
\(=\lim\limits_{x\to +\infty}[\frac{5}{\sqrt[3]{x^3+10x+\frac{25}{x}}+\sqrt[3]{x^2+5x}+x}-\frac{-3+\frac{6}{x}}{\sqrt{1-\frac{3}{x}+\frac{6}{x^2}}+1}]\)
\(=(0-\frac{-3}{2})=\frac{3}{2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(x-1\right)^2}{x\left(x^2+5\right)}=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-2x+1}{x\left(x^2+5\right)}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1-\dfrac{2}{x}+\dfrac{1}{x^2}}{x\left(1+\dfrac{5}{x^2}\right)}=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{x}\cdot\lim\limits_{x\rightarrow+\infty}\dfrac{1-\dfrac{2}{x}+\dfrac{1}{x^2}}{1+\dfrac{5}{x^2}}\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow+\infty}\dfrac{1}{x}=+\infty\\\lim\limits_{x\rightarrow+\infty}\dfrac{1-\dfrac{2}{x}+\dfrac{1}{x^2}}{1+\dfrac{5}{x^2}}=\dfrac{1}{1}=1>0\end{matrix}\right.\)
\(a=\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)\left(x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}\right)}{\left(x^3-3x-2\right)\left(x+\sqrt[]{x+2}\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+1\right)\left(x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}\right)}{\left(x-2\right)\left(x+1\right)^2\left(x+\sqrt[]{x+2}\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}}{\left(x+1\right)\left(x+\sqrt[]{x+2}\right)}=...\)
\(b=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[]{1+2x}-x-1\right)+\left(x+1-\sqrt[3]{1+3x}\right)}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x^2}{\sqrt[]{1+2x}+x+1}+\dfrac{x^3+3x^2}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt[]{1+2x}+x+1}+\dfrac{x+3}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{1+3x}+\sqrt[3]{\left(1+3x\right)^2}}\right)\)
\(=...\)
\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(\sqrt[]{5+4x}-2x-3\right)+\left(2x+3-\sqrt[3]{7+6x}\right)}{x^3+x^2-x-1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{5+4x-\left(2x+3\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(2x+3\right)^3-\left(7+6x\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4\left(x+1\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(x+1\right)^2\left(8x+20\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4}{2x+3+\sqrt[]{5+4x}}+\dfrac{8x+20}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{x-1}\)
\(=...\)
1.
Do \(\lim\limits_{x\rightarrow2}\left(3x-5\right)=1>0\)
\(\lim\limits_{x\rightarrow2}\left(x-2\right)^2=0\)
\(\left(x-2\right)^2>0;\forall x\ne2\)
\(\Rightarrow\lim\limits_{x\rightarrow2}\dfrac{3x-5}{\left(x-2\right)^2}=+\infty\)
2.
\(\lim\limits_{x\rightarrow1^-}\left(2x-7\right)=-5< 0\)
\(\lim\limits_{x\rightarrow1^-}\left(x-1\right)=0\)
\(x-1< 0;\forall x< 1\)
\(\Rightarrow\lim\limits_{x\rightarrow1^-}\dfrac{2x-7}{x-1}=+\infty\)
3.
\(\lim\limits_{x\rightarrow1^+}\left(2x-7\right)=-5< 0\)
\(\lim\limits_{x\rightarrow1^+}\left(x-1\right)=0\)
\(x-1>0;\forall x>1\)
\(\Rightarrow\lim\limits_{x\rightarrow1^+}\dfrac{2x-7}{x-1}=-\infty\)
\(lim_{x\rightarrow+\infty}\frac{x^5-3x}{x^2+1}=lim_{x\rightarrow+\infty}\frac{x^3-\frac{3}{x}}{1+\frac{1}{x^2}}=lim_{x\rightarrow+\infty}\frac{x^3}{1}=+\infty\)