a)

$\underset{x\to 2}{\lim }\frac{\sqrt{3x-2}-2}{x-2}$

zauważmy, że: 

$3x-2\ge 0\wedge x-2\ne 0$

$3x\ge 2x\ne 2$

$x\ge \frac{2}{3}$

$x\in ⟨\frac{2}{3},2)\cup \left(2,+\infty \right)$

zatem

$\underset{x\to 2}{\lim }\frac{\sqrt{3x-2}-2}{x-2}\cdot \frac{\sqrt{3x-2}+2}{\sqrt{3x-2}+2}=$

$=\underset{x\to 2}{\lim }\frac{\left|3x-2\right|-4}{\left(x-2\right)\left(\sqrt{3x-2}+2\right)}=$        

$=\underset{x\to 2}{\lim }\frac{3x-2-4}{\left(x-2\right)\left(\sqrt{3x-2}+2\right)}=$  

$=\underset{x\to 2}{\lim }\frac{3x-6}{\left(x-2\right)\left(\sqrt{3x-2}+2\right)}=$

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

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

$=\frac{3}{\sqrt{3\cdot 2-2}+2}=$  

$=\frac{3}{2+2}=\frac{3}{4}$  

---

b)

$\underset{x\to 0}{\lim }\frac{\sqrt{x+3}-\sqrt{2x+3}}{x+x^2}$

zauważmy, że:

1)

$x+3\ge 0$

$x\ge -3$

2)

$2x+3\ge 0$

$2x\ge -3$

$x\ge -\frac{3}{2}$

3)

$x+x^2\ne 0$

$x\left(1+x\right)\ne 0$

$x\ne 0\wedge x\ne -1$

wobec tego:

$x\in ⟨-\frac{3}{2},-1)\cup \left(-1,0\right)\cup \left(0,+\infty \right)$

zatem

$\underset{x\to 0}{\lim }\frac{\sqrt{x+3}-\sqrt{2x+3}}{x+x^2}\cdot \frac{\sqrt{x+3}+\sqrt{2x+3}}{\sqrt{x+3}+\sqrt{2x+3}}=$

$=\underset{x\to 0}{\lim }\frac{\left|x+3\right|-\left|2x+3\right|}{\left(x+x^2\right)\left(\sqrt{x+3}+\sqrt{2x+3}\right)}=$    

$=\underset{x\to 0}{\lim }\frac{x+3-2x-3}{x\left(1+x\right)\left(\sqrt{x+3}+\sqrt{2x+3}\right)}=$           

$=\underset{x\to 0}{\lim }\frac{-x}{x\left(1+x\right)\left(\sqrt{x+3}+\sqrt{2x+3}\right)}=$  

$=\underset{x\to 0}{\lim }\frac{-1}{\left(1+x\right)\left(\sqrt{x+3}+\sqrt{2x+3}\right)}=$  

$=\frac{-1}{\left(1+0\right)\left(\sqrt{0+3}+\sqrt{2\cdot 0+3}\right)}=$  

$=\frac{-1}{1\cdot \left(\sqrt{3}+\sqrt{3}\right)}=$  

$=\frac{-1}{2\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=$

$=-\frac{\sqrt{3}}{6}$   

---

c)

$\underset{x\to 5}{\lim }\frac{\sqrt{x-1}-2}{x-5}$

zauważmy, że:

1)

$x-1\ge 0$    

$x\ge 1$

2)

$x-5\ne 0$

$x\ne 5$

wobec tego:

$x\in ⟨1,5)\cup \left(5,+\infty \right)$

zatem

$\underset{x\to 5}{\lim }\frac{\sqrt{x-1}-2}{x-5}\cdot \frac{\sqrt{x-1}+2}{\sqrt{x-1}+2}=$

$=\underset{x\to 5}{\lim }\frac{\left|x-1\right|-4}{\left(x-5\right)\left(\sqrt{x-1}+2\right)}=$       

$=\underset{x\to 5}{\lim }\frac{x-5}{\left(x-5\right)\left(\sqrt{x-1}+2\right)}=$

$=\underset{x\to 5}{\lim }\frac{1}{\sqrt{x-1}+2}=$

$=\frac{1}{\sqrt{5-1}+2}=$

$=\frac{1}{2+2}=\frac{1}{4}$     

---

d)

$\underset{x\to 1}{\lim }\frac{\sqrt{x+3}-2}{x-1}$

zauważmy, że:

1)

$x+3\ge 0$

$x\ge -3$

2)

$x-1\ne 0$

$x\ne 1$

wobec tego:

$x\in ⟨-3,1)\cup \left(1,+\infty \right)$

zatem

$\underset{x\to 1}{\lim }\frac{\sqrt{x+3}-2}{x-1}\cdot \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}=$

$=\underset{x\to 1}{\lim }\frac{\left|x+3\right|-4}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=$         

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

$=\underset{x\to 1}{\lim }\frac{1}{\sqrt{x+3}+2}=$

$=\frac{1}{\sqrt{1+3}+2}=\frac{1}{4}$    

---

e)

$\underset{x\to 4}{\lim }\frac{\sqrt{1+2x}-3}{\sqrt{x}-2}$

zauważmy, że:

1)

 1+2x>=0 

$2x\ge -1$

$x\ge -\frac{1}{2}$

2)

$x\ge 0$

3)

$\sqrt{x}-2\ne 0$

$\sqrt{x}\ne 2$

$x\ne 4$

wobec tego:

$x\in ⟨0,4)\cup \left(4,+\infty \right)$

zatem

$\underset{x\to 4}{\lim }\frac{\sqrt{1+2x}-3}{\sqrt{x}-2}\cdot \frac{\sqrt{x}+2}{\sqrt{x}+2}=$

$=\underset{x\to 4}{\lim }\frac{\left(\sqrt{1+2x}-3\right)\left(\sqrt{x}+2\right)}{x-4}\cdot \frac{\sqrt{1+2x}+3}{\sqrt{1+2x}+3}=$

$=\underset{x\to 4}{\lim }\frac{\left(1+2x-9\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{1+2x}+3\right)}=$             

$=\underset{x\to 4}{\lim }\frac{\left(2x-8\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{1+2x}+3\right)}=$

$=\underset{x\to 4}{\lim }\frac{2\left(x-4\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{1+2x}+3\right)}=$

$=\underset{x\to 4}{\lim }\frac{2\left(\sqrt{x}+2\right)}{\sqrt{1+2x}+3}=$    

$=\frac{2\left(\sqrt{4}+2\right)}{\sqrt{1+2\cdot 4}+3}=$

$=\frac{2\cdot 4}{3+3}=\frac{8}{6}=\frac{4}{3}$   

---

f)

$\underset{x\to 3}{\lim }\frac{\sqrt{x+13}-2\sqrt{x+1}}{x^2-9}$

zauważmy, że:

1)

$x+13\ge 0$   

$x\ge -13$

2)

$x+1\ge 0$   

$x\ge -1$

3)

$x^2-9\ne 0$

$x^2\ne 9$

$x\ne 3\wedge x\ne -3$

wobec tego:

$x\in ⟨-1,3)\cup \left(3,+\infty \right)$

zatem

$\underset{x\to 3}{\lim }\frac{\sqrt{x+13}-2\sqrt{x+1}}{x^2-9}\cdot \frac{\sqrt{x+13}+2\sqrt{x+1}}{\sqrt{x+13}+2\sqrt{x+1}}=$       

$=\underset{x\to 3}{\lim }\frac{\left|x+13\right|-4\left|x+1\right|}{\left(x^2-9\right)\left(\sqrt{x+13}+2\sqrt{x+1}\right)}=$  

$=\underset{x\to 3}{\lim }\frac{x+13-4\left(x+1\right)}{\left(x^2-9\right)\left(\sqrt{x+13}+2\sqrt{x+1}\right)}=$  

$=\underset{x\to 3}{\lim }\frac{x+13-4x-4}{\left(x^2-9\right)\left(\sqrt{x+13}+2\sqrt{x+1}\right)}=$  

$=\underset{x\to 3}{\lim }\frac{-3x+9}{\left(x-3\right)\left(x+3\right)\left(\sqrt{x+13}+2\sqrt{x+1}\right)}=$

$=\underset{x\to 3}{\lim }\frac{-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(\sqrt{x+13}+2\sqrt{x+1}\right)}=$   

$=\underset{x\to 3}{\lim }\frac{-3}{\left(x+3\right)\left(\sqrt{x+13}+2\sqrt{x+1}\right)}=$  

$=\frac{-3}{\left(3+3\right)\left(\sqrt{3+13}+2\sqrt{3+1}\right)}=$  

$=\frac{-3}{6\left(4+4\right)}=\frac{-3}{48}=-\frac{1}{16}$