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1/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}.\sqrt[4]{1+8x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}-1}{x}\)
Liên hợp dài quá ko muốn gõ tiếp, bạn tự đặt nhân tử chung rồi liên hợp nhé, kết quả ra 5
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{1+7x}-2-\left(x^3-3x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{7\left(x-1\right)}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)^2\left(x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{7}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)\left(x+2\right)=\dfrac{7}{12}\)
3/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x^3-x^2+1}{2x^2+3x-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x-1+\dfrac{1}{x^2}}{2+\dfrac{3}{x}-\dfrac{1}{x^2}}=-\infty\)
4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{4x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{1+\dfrac{1}{\sqrt[6]{x}}+\dfrac{1}{\sqrt[4]{x}}}{\sqrt{4+\dfrac{1}{x}}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}\)
5/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{1+\dfrac{2}{x^2}}}{\sqrt[3]{8+\dfrac{1}{x}+\dfrac{1}{x^3}}}=\dfrac{1-1}{\sqrt[3]{8}}=0\)
6/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+3x-7}}{\sqrt[3]{27x^3+5x^2+x-4}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4+\dfrac{3}{x}-\dfrac{7}{x^2}}}{\sqrt[3]{27+\dfrac{5}{x}+\dfrac{1}{x^2}-\dfrac{4}{x^3}}}=\dfrac{-\sqrt{4}}{\sqrt[3]{27}}=\dfrac{-2}{3}\)
1/ \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-2+2-\sqrt[3]{8-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2x}{\sqrt{1+x}+1}+\dfrac{x}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+x}+1}+\dfrac{1}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\right)=\dfrac{13}{12}\)
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{x+3}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-2-\left(\sqrt{x+3}-2\right)}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{x-1}{\sqrt{x+3}+2}}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{1}{\sqrt{x+3}+2}}{x-2}=\dfrac{1}{6}\)
3/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\sqrt{5-x^2}}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-2+2-\sqrt{5-x^2}}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{\left(x^2-1\right)}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{x^2-1}{2+\sqrt{5-x^2}}}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\dfrac{1}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{1}{2+\sqrt{5-x^2}}\right)=\dfrac{1}{3}\)
4/ \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-\sqrt[3]{8x+43}}{2x^2+3x-2}=\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-3-\left(\sqrt[3]{8x+43}-3\right)}{\left(2x-1\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{x+2}{\sqrt{x+11}+3}-\dfrac{8\left(x+2\right)}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{\left(2x-1\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{1}{\sqrt{x+11}+3}-\dfrac{8}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{2x-1}=\dfrac{7}{270}\)
5/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-\sqrt[m]{1+bx}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-1-\left(\sqrt[m]{1+bx}-1\right)}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{ax}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{bx}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{a}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{b}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}\)
\(=\dfrac{a}{n}-\dfrac{b}{m}\)
6/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-\sqrt{1+4x}+\sqrt{1+4x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\left(\sqrt[3]{1+6x}-1\right)+\sqrt{1+4x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\dfrac{6x}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4x}{\sqrt{1+4x}+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{6\sqrt{1+4x}}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4}{\sqrt{1+4x}+1}\right)=4\)
1.
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{x^2-3x-4}}{1-x^2}=\lim\limits_{x\to (-1)-}\frac{\sqrt{(x+1)(x-4)}}{(1-x)(1+x)}\)
\(=\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{(x-1)\sqrt{-(x+1)}}=-\infty\) do:
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{x-1}=\frac{-\sqrt{5}}{2}<0\) và \(\lim\limits_{x\to (-1)-}\frac{1}{\sqrt{-(x+1)}}=+\infty\)
2.
\(\lim\limits_{x\to 2+}\left(\frac{1}{x-2}-\frac{x+1}{\sqrt{x+2}-2}\right)=\lim\limits_{x\to 2+}\frac{1-(x+1)(\sqrt{x+2}+2)}{x-2}=-\infty\) do:
\(\lim\limits_{x\to 2+}\frac{1}{x-2}=+\infty\) và \(\lim\limits_{x\to 2+}[1-(x+1)(\sqrt{x+2}+2)]=-11<0\)
a) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}=\lim\limits_{x\rightarrow0}\frac{2x}{2x\left(\sqrt{1+2x}+1\right)}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{1+2x}+1}=\frac{1}{2}\)
b) \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}=\lim\limits_{x\rightarrow0}\frac{4x\left(\sqrt{9+x}+3\right)}{x}=\lim\limits_{x\rightarrow0}[4\left(\sqrt{9+x}+3\right)=24\)
c) \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{x-2}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\frac{1}{\sqrt{x+7}+3}=\frac{1}{6}\)
d) \(\lim\limits_{x\rightarrow1}\frac{3x-2-\sqrt{4x^2-x-2}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\frac{\left(3x-2\right)^2-\left(4x^2-4x-2\right)}{(x^2-3x+2)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(5x-6\right)}{\left(x-1\right)\left(x-2\right)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\frac{1}{2}\\ \\\\ \\ \\ \\ \)
e)\(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\frac{2x+7-\left(x^2-8x+16\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x-9\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{x-9}{\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=-8\)
f) \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}-3}{2-\sqrt{x+3}}=\lim\limits_{x\rightarrow1}\frac{(2x-2)\left(2+\sqrt{x+3}\right)}{\left(1-x\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\frac{-2\left(2+\sqrt{x+3}\right)}{\sqrt{2x+7}+3}=\frac{-4}{3}\)
g) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}=\lim\limits_{x\rightarrow0}\frac{x^2\left(\sqrt{x^2+16}+4\right)}{x^2\left(\sqrt{x^2+1}+1\right)}=4\)
h)
\(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-3}{x-4}+\lim\limits_{x\rightarrow4}\frac{3-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{1}{\sqrt{x+5}+4}+\lim\limits_{x\rightarrow4}\frac{8-2x}{\left(x-4\right)\left(3+\sqrt{2x+1}\right)}=\frac{1}{7}-\frac{1}{3}=\frac{-4}{21}\)
k) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{x+4}-2}{x}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+1}+1}+\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
\(\lim\limits_{x\rightarrow1}\dfrac{x^3-3x+2}{x^4-4x+3}=\lim\limits_{x\rightarrow1}\dfrac{\left(x+2\right)\left(x-1\right)^2}{\left(x^2+2x+3\right)\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\dfrac{x+2}{x^2+2x+3}=\dfrac{1}{2}\)
\(\lim\limits_{x\rightarrow2^-}\dfrac{x^3+x^2-4x-4}{x^2-4x+4}=\lim\limits_{x\rightarrow2^-}\dfrac{\left(x-2\right)\left(x^2+3x+2\right)}{\left(x-2\right)^2}=\lim\limits_{x\rightarrow2^-}\dfrac{x^2+3x+2}{x-2}=-\infty\)
\(\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)^{20}}{\left(x^3-12x+16\right)^{10}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}\left(x-2\right)^{20}}{\left(x+4\right)^{10}\left(x-2\right)^{20}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}}{\left(x+4\right)^{10}}=\dfrac{3^{10}}{2^{10}}\)
\(\lim\limits_{x\rightarrow0^-}\dfrac{4x^2+5x}{x^2}=\lim\limits_{x\rightarrow0^-}\dfrac{4x+5}{x}=-\infty\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+2}-1}{\sqrt{x+5}-2}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(\sqrt{x+5}+2\right)}{\left(x+1\right)\left(\sqrt{x+2}+1\right)}=\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+5}+2}{\sqrt{x+2}+1}=2\)
Câu 1:
\(\lim _{x\to +\infty}(2x-1-\sqrt{4x^2-4x-3})=\lim_{x\to +\infty}\frac{(2x-1)^2-(4x^2-4x-3)}{2x-1+\sqrt{4x^2-4x-3}}\) (liên hợp)
\(=\lim_{x\to +\infty}\frac{4}{2x-1+\sqrt{4x^2-4x-3}}=4\lim_{x\to +\infty}\frac{1}{2x-1+\sqrt{4x^2-4x-3}}\)
Ta thấy với \(x\to +\infty\Rightarrow 2x-1+\sqrt{4x^2-4x-3}\to +\infty\)
Do đó: \(\lim_{x\to +\infty}\frac{1}{2x-1+\sqrt{4x^2-4x-3}}=0\) (theo dạng \(\lim _{t\to \infty}\frac{1}{t}=0\) )
\(\Rightarrow \lim _{x\to +\infty}(2x-1-\sqrt{4x^2-4x-3})=0\)
Câu 3:
\(\lim_{x\to 1+} (x^3-1)\sqrt{\frac{x}{x^2-1}}=\lim_{x\to 1+}(x^2+x+1)\sqrt{\frac{x(x-1)^2}{x^2-1}}\)
\(=\lim_{x\to 1+}(x^2+x+1)\sqrt{\frac{x(x-1)}{x+1}}=(1+1+1)\sqrt{\frac{1.0}{1+1}}=0\)
Câu 2:
\(\lim_{x\to 3}\frac{\sqrt{2x^2-2}-\sqrt{4x-3}+2x-7}{9-x^2}=\lim_{x\to 3}\frac{\sqrt{2x^2-2}-4}{9-x^2}-\lim_{x\to 3}\frac{\sqrt{4x-3}-3}{9-x^2}+\lim_{x\to 3}\frac{2x-6}{9-x^2}\)
Ta có:
\(\lim_{x\to 3}\frac{2x^2-2-16}{(\sqrt{2x^2-2}+4)(9-x^2)}=\lim_{x\to 3}\frac{2(x^2-9)}{(\sqrt{2x^2-2}+4)(9-x^2)}=\lim_{x\to 3}\frac{-2}{\sqrt{2x^2-2}+4}=\frac{-1}{4}\) (1)
\(\lim_{x\to 3}\frac{\sqrt{4x-3}-3}{9-x^2}=\lim_{x\to 3}\frac{4x-3-9}{(\sqrt{4x-3}+3)(9-x^2)}=\lim_{x\to 3}\frac{4(x-3)}{(\sqrt{4x-3}+3)(9-x^2)}\)
\(=\lim_{x\to 3}\frac{-4}{(\sqrt{4x-3}+3)(3+x)}=-\frac{1}{9}\) (2)
\(\lim _{x\to 3}\frac{2x-6}{9-x^2}=\lim_{x\to 3}\frac{2(x-3)}{9-x^2}=\lim_{x\to 3}\frac{-2}{x+3}=\frac{-1}{3}\) (3)
Từ \((1); (2); (3)\Rightarrow \lim_{x\to 3}\frac{\sqrt{2x^2-2}-\sqrt{4x-3}+2x-7}{9-x^2}=\frac{-1}{4}+\frac{1}{9}-\frac{1}{3}=\frac{-17}{36}\)
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/ \(=\lim\limits_{x\rightarrow-2}\dfrac{2x+5-1}{\left(x-2\right)\left(x+2\right)\left(\sqrt{2x+5}+1\right)}=\lim\limits_{x\rightarrow-2}\dfrac{2}{\left(x-2\right)\left(\sqrt{2x+5}+1\right)}=\dfrac{2}{\left(-2-2\right)\left(\sqrt[2]{2.\left(-2\right)+5}+1\right)}=\dfrac{2}{\left(-4\right).2}=-\dfrac{1}{4}\)