$\text{a)}$  

$\underset{x\to -3}{\lim }\sqrt{x^2-2x+3}=\sqrt{{\left(-3\right)}^2-2\cdot \left(-3\right)+3}=\sqrt{9+6+3}=\sqrt{18}=3\sqrt{2}$  

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$\text{b)}$     

$\underset{x\to -1}{\lim }\sqrt{\frac{2}{3}{x}^3-3x+\frac{1}{3}}=\sqrt{\frac{2}{3}\cdot {\left(-1\right)}^3-3\cdot \left(-1\right)+\frac{1}{3}}=\sqrt{-\frac{2}{3}+3+\frac{1}{3}}=$  

$=\sqrt{2\frac{2}{3}}=\sqrt{\frac{8}{3}}=\frac{2\sqrt{2}}{\sqrt{3}}=\frac{2\sqrt{6}}{3}$  

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$\text{c)}$  

$\underset{x\to -2}{\lim }\sqrt[3]{4x^3-x^2+4}=\sqrt[3]{4\cdot {\left(-2\right)}^3-{\left(-2\right)}^2+4}=\sqrt[3]{4\cdot \left(-8\right)-4+4}=$  

$=\sqrt[3]{-32}=\sqrt[3]{-8\cdot 4}=-2\sqrt[3]{4}$  

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$\text{d)}$  

$\underset{x\to 1}{\lim }\sqrt[3]{2\left|x-2\right|-\left|1-3x\right|}=\sqrt[3]{2\left|1-2\right|-\left|1-3\right|}=\sqrt[3]{2\left|-1\right|-\left|-2\right|}=\sqrt[3]{2\cdot 1-2}=\sqrt[3]{0}=0$  

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$\text{e)}$  

$\underset{x\to 2}{\lim }\frac{x-2}{\sqrt{x-1}-1}=\underset{x\to 2}{\lim }\frac{\left(x-2\right)\cdot \left(\sqrt{x-1}+1\right)}{\left(\sqrt{x-1}-1\right)\left(\sqrt{x-1}+1\right)}=$  

$=\underset{x\to 2}{\lim }\frac{\left(x-2\right)\cdot \left(\sqrt{x-1}+1\right)}{{\sqrt{x-1}}^2-1^2}=\underset{x\to 2}{\lim }\frac{\left(x-2\right)\cdot \left(\sqrt{x-1}+1\right)}{\left|x-1\right|-1}=$  

$=\underset{x\to 2}{\lim }\frac{\left(x-2\right)\cdot \left(\sqrt{x-1}+1\right)}{x-1-1}=\underset{x\to 2}{\lim }\frac{\left(x-2\right)\cdot \left(\sqrt{x-1}+1\right)}{x-2}=$  

$=\underset{x\to 2}{\lim }\left(\sqrt{x-1}+1\right)=\sqrt{2-1}+1=\sqrt{1}+1=1+1=2$  

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$\text{f)}$  

$\underset{x\to 0}{\lim }\frac{\sqrt{x+1}-\sqrt{1-x}}{2x}=\underset{x\to 0}{\lim }\frac{\sqrt{x+1}-\sqrt{1-x}}{2x}\cdot \frac{\sqrt{x+1}+\sqrt{1-x}}{\sqrt{x+1}+\sqrt{1-x}}=$   

$=\underset{x\to 0}{\lim }\frac{{\sqrt{x+1}}^2-{\sqrt{1-x}}^2}{2x\left(\sqrt{x+1}+\sqrt{1-x}\right)}=\underset{x\to 0}{\lim }\frac{\left|x+1\right|-\left|1-x\right|}{2x\left(\sqrt{x+1}+\sqrt{1-x}\right)}=$  

$=\underset{x\to 0}{\lim }\frac{x+1-\left(1-x\right)}{2x\left(\sqrt{x+1}+\sqrt{1-x}\right)}=\underset{x\to 0}{\lim }\frac{x+1-1+x}{2x\left(\sqrt{x+1}+\sqrt{1-x}\right)}=$  

$=\underset{x\to 0}{\lim }\frac{2x}{2x\left(\sqrt{x+1}+\sqrt{1-x}\right)}=\underset{x\to 0}{\lim }\frac{1}{\sqrt{x+1}+\sqrt{1-x}}=$  

$=\frac{1}{\sqrt{0+1}+\sqrt{1-0}}=\frac{1}{1+1}=\frac{1}{2}$