Bài 2:

\(\lim\limits_{x\to 2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}=\lim\limits_{x\to 2}\frac{x^2-x-2}{(x+\sqrt{x+2}).\frac{4x+1-9}{\sqrt{4x+1}+3}}=\lim\limits_{x\to 2}\frac{(x-2)(x+1)(\sqrt{4x+1}+3)}{(x+\sqrt{x+2}).4(x-2)}=\lim\limits_{x\to 2}\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}=\frac{9}{8}\)

Bài 3:

\(\lim\limits_{x\to 0-}\frac{1-\sqrt[3]{x-1}}{x}=-\infty \)

\(\lim\limits_{x\to 0+}\frac{1-\sqrt[3]{x-1}}{x}=+\infty \)

Bài 4:

\(\lim\limits_{x\to -\infty}\frac{x^2-5x+1}{x^2-2}=\lim\limits_{x\to -\infty}\frac{1-\frac{5}{x}+\frac{1}{x^2}}{1-\frac{2}{x^2}}=1\)

Bài 5:

\(\lim\limits_{x\to +\infty}\frac{2x^2-4}{x^3+3x^2-9}=\lim\limits_{x\to +\infty}\frac{\frac{2}{x}-\frac{4}{x^3}}{1+\frac{3}{x}-\frac{9}{x^3}}=0\)

Bài 6:

\(\lim\limits_{x\to 2- }\frac{2x-1}{x-2}=\lim\limits_{x\to 2-}\frac{2(x-2)+3}{x-2}=\lim\limits_{x\to 2-}\left(2+\frac{3}{x-2}\right)=-\infty \)

Bài 7:

\(\lim\limits _{x\to 3+ }\frac{8+x-x^2}{x-3}=\lim\limits _{x\to 3+}\frac{1}{x-3}.\lim\limits _{x\to 3+}(8+x-x^2)=2(+\infty)=+\infty \)

Bài 8:

\(\lim\limits _{x\to -\infty}(8+4x-x^3)=\lim\limits _{x\to -\infty}(-x^3)=+\infty \)

Bài 9:

\(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\)

\(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)

Bài 1:

\(\lim\limits_{x\to1+}\frac{2x^2-3x+1}{x^3-x^2-x+1}=\lim\limits_{x\to1+}\frac{\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=\lim\limits_{x\to1+}\frac{2x-1}{x^2-1}\)

\(\lim\limits_{x\to 1+}\frac{1}{x^2-1}.\lim\limits_{x\to 1+}(2x-1)=1.(+\infty)=+\infty \)

Tương tự \(\lim\limits_{x\to 1-} \frac{2x^2-3x+1}{x^3-x^2-x+1}=-\infty \)

a: \(\lim\limits_{x\rightarrow2}\dfrac{1-\sqrt{x^2+3}}{-x^2+3x-2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x^2+3}-1}{x^2-3x+2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{2^2+3}-1}{2^2-3\cdot2+2}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\sqrt{2^2+3}-1=\sqrt{7}-1>0\\\lim\limits_{x\rightarrow2}2^2-3\cdot2+2=0\end{matrix}\right.\)

b: \(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{4x-1}+3}{x^2-4}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{4x-1-9}{\sqrt{4x-1}-3}\cdot\dfrac{1}{x^2-4}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}\cdot\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}=\dfrac{4\cdot2-10}{\sqrt{4\cdot2-1}-3}=\dfrac{-2}{\sqrt{7}-3}>0\\\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{1}{\left(2+2\right)\cdot\left(2-2\right)}=+\infty\end{matrix}\right.\)

1/ \(=\lim\limits_{x\rightarrow0}\dfrac{3\left(1+3x\right)^2.3+4.4\left(1-4x\right)^3}{1}=...\left(thay-x-vo\right)\)

2/ \(=\lim\limits_{x\rightarrow2}\dfrac{2.2.x-5}{3x^2-3}=\dfrac{4.2-5}{3.4-3}=\dfrac{1}{3}\)

3/ \(=\lim\limits_{x\rightarrow1}\dfrac{4x^3-3}{3x^2+2}=\dfrac{4.1-3}{3.1-2}=1\)

Xai L'Hospital nhe :v

a: \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+8}-4}{x-4}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2x+8-16}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2\left(x-4\right)}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2}{\sqrt{2x+8}+4}=\dfrac{2}{\sqrt{2\cdot4+8}+4}\)

\(=\dfrac{2}{\sqrt{8+8}+4}=\dfrac{2}{4+4}=\dfrac{2}{8}=\dfrac{1}{4}\)

b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{\sqrt{4x+1}-3}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{\dfrac{4x+1-9}{\sqrt{4x+1}+3}}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{4\left(x-2\right)}\cdot\left(\sqrt{4x+1}+3\right)\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}{4}\)

\(=\dfrac{\left(2+2\right)\left(\sqrt{4\cdot2+1}+3\right)}{4}=\sqrt{9}+3=6\)

c: \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-\sqrt{x+2}}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\dfrac{4-x-2}{2+\sqrt{x+2}}}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-x}\cdot\left(\sqrt{x+2}+2\right)\)

\(=\lim\limits_{x\rightarrow2}\left(-\sqrt{x+2}-2\right)\)

\(=-\sqrt{2+2}-2=-2-2=-4\)

\(a=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{9+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\)

\(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{9+\sqrt{x+7}}}{x-1}=\frac{29}{36}\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(2-\frac{3}{x}\right)^2.x^3\left(4+\frac{7}{x}\right)^3}{x^3\left(3+\frac{1}{x^3}\right).x^2\left(10+\frac{9}{x^2}\right)}=\frac{2.4}{3.10}=\frac{4}{15}\)

\(c=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}-\left(2x+1\right)+\left(2x+1-\sqrt[3]{1+6x}\right)}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\frac{\frac{-4x^2}{\sqrt{1+4x}+2x+1}+\frac{8x^3+12x^2}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(\frac{-4}{\sqrt{1+4x}+2x+1}+\frac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}\right)=\frac{-4}{1+1}+\frac{12}{1+1+1}=2\)

\(d=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+6x}\left(\sqrt{1+4x}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{1+6x}-1}{x}\)

\(=\lim\limits_{x\rightarrow0}\frac{4x\sqrt{1+6x}}{x\left(\sqrt{1+4x}+1\right)}+\lim\limits_{x\rightarrow0}\frac{6x}{x\left(\sqrt{1+6x}+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\frac{4\sqrt{1+6x}}{\sqrt{1+4x}+1}+\lim\limits_{x\rightarrow0}\frac{6}{\sqrt{1+6x}+1}=\frac{4}{1+1}+\frac{6}{1+1}=5\)

\(e=\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+4x}\left(\sqrt{1+2x}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+4x}-1}{x}\)

\(=\lim\limits_{x\rightarrow0}\frac{2x\sqrt[3]{1+4x}}{x\left(\sqrt{1+2x}+1\right)}+\lim\limits_{x\rightarrow0}\frac{4x}{x\left(\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\frac{2\sqrt[3]{1+4x}}{\sqrt{1+2x}+1}+\lim\limits_{x\rightarrow0}\frac{4}{\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1}=\frac{2}{1+1}+\frac{4}{1+1+1}=\frac{7}{3}\)

a/ L'Hospital:

 \(=\lim\limits_{x\rightarrow2}\dfrac{x-\left(x+2\right)^{\dfrac{1}{2}}}{\left(4x+1\right)^{\dfrac{1}{2}}-3}=\lim\limits_{x\rightarrow2}\dfrac{1-\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}{\dfrac{1}{2}\left(4x+1\right)^{-\dfrac{1}{2}}.4}=\dfrac{1-\dfrac{1}{2}.4^{-\dfrac{1}{2}}}{2.9^{-\dfrac{1}{2}}}=\dfrac{9}{8}\)

b/ L'Hospital:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)

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) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{x+4}-2}{x}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+1}+1}+\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)

Em kiểm tra lại đề, chỗ \(f\left(x\right)-32\) kia có vẻ sai, vì như thế thì biểu thức đã cho ko phải dạng vô định

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:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)

\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)