a: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{4+\dfrac{3}{x}}{2}=\dfrac{4}{2}=2\)

b: \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}}{3+\dfrac{1}{x}}=0\)

c: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}}{1+\dfrac{1}{x}}=1\)

a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right) = \mathop {\lim }\limits_{x \to 2} {x^2} - \mathop {\lim }\limits_{x \to 2} \left( {4x} \right) + 3 = {2^2} - 4.2 + 3 =  - 1\)

b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \left( {x - 2} \right) = \mathop {\lim }\limits_{x \to 3} x - 2 = 3 - 2 = 1\)

c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x  + 1}} = \frac{1}{{\sqrt 1  + 1}} = \frac{1}{2}\)

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

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

\(=-1+\dfrac{3}{3}=0\)

\(a=\lim\limits_{x\rightarrow0}\frac{x^2}{x\left(\sqrt{1+x^2}+1\right)}=\lim\limits_{x\rightarrow0}\frac{x}{\sqrt{1+x^2}+1}=\frac{0}{2}=0\)

\(b=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}=\lim\limits_{x\rightarrow1}\frac{\frac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{\left(x-1\right)\left(x+1\right)}{2+\sqrt{5-x^2}}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\left(\frac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{x+1}{2+\sqrt{5-x^2}}\right)=\frac{1}{12}+\frac{1}{2}=\frac{7}{12}\)

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

\(d=\frac{\sqrt[3]{6}}{0}=+\infty\)

\(lim_{x\rightarrow1}\frac{x^3+2x-3}{x^2-x}\)   

\(=lim_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2+x+3\right)}{x\left(x-1\right)}\)   

\(=lim_{x\rightarrow1}\frac{x^2+x+3}{x}\)   

\(=\frac{1^2+1+3}{1}\)   

\(=5\)   

\(lim_{x\rightarrow1}\frac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}\)   

\(=lim_{x\rightarrow1}\frac{\left(2x+2\right)-\left(3x+1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{2x+2-3x-1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-x+1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-1\left(x-1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-1}{\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=\frac{-1}{\sqrt{2\cdot1+2}+\sqrt{3\cdot1+1}}\)   

\(=\frac{-1}{2+2}=\frac{-1}{4}\)

https://drive.google.com/file/d/14Q-YI3szy-rePnIHWGD35RKCWiCXCT6k/view?usp=sharing

https://drive.google.com/file/d/1425SNt8hu4qt2y1kIcnhIvcxPfODsY1T/view?usp=sharing

a) Ta có \(t = \frac{1}{x},\) nên khi x tiến đến 0 thì t tiến đến dương vô cùng do đó

\(\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}} = \mathop {\lim }\limits_{t \to  + \infty } {\left( {1 + \frac{1}{t}} \right)^t} = e\)

b) \(\ln y = \ln {\left( {1 + x} \right)^{\frac{1}{x}}} = \frac{1}{x}\ln \left( {1 + x} \right)\)

\(\mathop {\lim }\limits_{x \to 0} \ln y = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = 1\)

c) \(t = {e^x} - 1 \Leftrightarrow {e^x} = t + 1 \Leftrightarrow x = \ln \left( {t + 1} \right)\)

\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{t \to 0} \frac{t}{{\ln \left( {t + 1} \right)}} = 1\)

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

\(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4+4x}-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4+4x}-2+2-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x+4-4}{\sqrt{4x+4}+2}+\dfrac{8-8+x}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt{4x+4}+2}+\dfrac{x}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)

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

\(=\dfrac{4}{\sqrt{4\cdot0+4}+2}+\dfrac{1}{4+2\cdot\sqrt[3]{8-0}+\sqrt[3]{\left(8-0\right)^2}}\)

\(=\dfrac{4}{2+2}+\dfrac{1}{4+2\cdot2+4}\)

\(=1+\dfrac{1}{12}=\dfrac{13}{12}\)