a)

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

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

$=\underset{x\to 0}{\lim }\frac{1+3x+3x+9x^2+2x^2+6x^3-1}{x}=$

$=\underset{x\to 0}{\lim }\frac{6x^3+11x^2+6x}{x}=$

$=\underset{x\to 0}{\lim }\frac{x\left(6x^2+11x+6\right)}{x}=$

$=\underset{x\to 0}{\lim }\left(6x^2+11x+6\right)=$

$=6\cdot 0^2+11\cdot 0+6=6$        

---

b)

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

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

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

$=\frac{3+0}{0+1}=3$     

---

c)

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

$=\underset{x\to 0}{\lim }\frac{1+6x+9x^2-\left(1+6x+12x^2+8x^3\right)}{x^2}=$

$=\underset{x\to 0}{\lim }\frac{1+6x+9x^2-x-6x-12x^2-8x^3}{x^2}=$

$=\underset{x\to 0}{\lim }\frac{-3x^2-8x^3}{x^2}=$

$=\underset{x\to 0}{\lim }\frac{x^2\left(-3-8x\right)}{x^2}=$

$=\frac{-3-8\cdot 0}{1}=-3$       

---

d)

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

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

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

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

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

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

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

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

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

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

$=\frac{1+2}{1^2+2\cdot 1+3}=\frac{3}{6}=\frac{1}{2}$  

---

e)

$\underset{x\to 1}{\lim }\frac{x^3-2x+1}{x^5-2x+1}=$

$=\underset{x\to 1}{\lim }\frac{x^3-x-x+1}{x^5-x-x+1}=$

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

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

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

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

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

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

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

$=\frac{1^2+1-1}{1^4+1^3+1^2+1-1}=$

$=\frac{1}{1+1+1+1-1}=\frac{1}{3}$   

---

f)

$\underset{x\to 1}{\lim }\frac{x+x^2+x^3+x^4+x^5+x^6+x^7-7}{x-1}=$  

$=\underset{x\to 1}{\lim }\frac{x^7-x^6+2x^6-2x^5+3x^5-3x^4+4x^4-4x^3+5x^3-5x^2+6x^2-6x+7x-7}{x-1}=$

$=\underset{x\to 1}{\lim }\frac{x^6\left(x-1\right)+2x^5\left(x-1\right)+3x^4\left(x-1\right)+4x^3\left(x-1\right)+5x^2\left(x-1\right)+6x\left(x-1\right)+7\left(x-1\right)}{x-1}=$

$=\underset{x\to 1}{\lim }\frac{\left(x-1\right)\left(x^6+2x^5+3x^4+4x^3+5x^2+6x+7\right)}{x-1}=$

$=\underset{x\to 1}{\lim }\left(x^6+2x^5+3x^4+4x^3+5x^2+6x+7\right)=$

$=1^6+2\cdot 1^5+3\cdot 1^4+4\cdot 1^3+5\cdot 1^2+6\cdot 1+7=$

$=1+2+3+4+5+6+7=28$       

---

g)

$\underset{x\to 2}{\lim }\frac{{\left(x^2-x-2\right)}^{20}}{{\left(x^3-12x+16\right)}^{10}}=$

$=\underset{x\to 2}{\lim }\frac{{\left(x^2+x-2x-2\right)}^{20}}{{\left(x^3-4x-8x+16\right)}^{10}}=$   

$=\underset{x\to 2}{\lim }\frac{{\left(x\left(x+1\right)-2\left(x+1\right)\right)}^{20}}{{\left(x\left(x^2-4\right)-8\left(x-2\right)\right)}^{10}}=$

$=\underset{x\to 2}{\lim }\frac{{\left(\left(x+1\right)\left(x-2\right)\right)}^{20}}{{\left(x\left(x-2\right)\left(x+2\right)-8\left(x-2\right)\right)}^{10}}=$   

$=\underset{x\to 2}{\lim }\frac{{\left(\left(x+1\right)\left(x-2\right)\right)}^{20}}{{\left(\left(x-2\right)\left(x\left(x+2\right)-8\right)\right)}^{10}}=$

$=\underset{x\to 2}{\lim }\frac{{\left(x+1\right)}^{20}\cdot {\left(x-2\right)}^{20}}{{\left(x-2\right)}^{10}\cdot {\left(x^2+2x-8\right)}^{10}}$

Zauważmy, że:

$x^2+2x-8=0$

$\Delta =2^2-4\cdot 1\cdot \left(-8\right)=4+32=36,\sqrt{\Delta }=6$

$x=\frac{-2-6}{2}=-4\text{lub}x=\frac{-2+6}{2}=2$

$x^2+2x-8=\left(x+4\right)\left(x-2\right)$       

$=\underset{x\to 2}{\lim }\frac{{\left(x+1\right)}^{20}\cdot {\left(x-2\right)}^{20}}{{\left(x-2\right)}^{10}\cdot {\left(\left(x+4\right)\left(x-2\right)\right)}^{10}}=$  

$=\underset{x\to 2}{\lim }\frac{{\left(x+1\right)}^{20}\cdot {\left(x-2\right)}^{20}}{{\left(x-2\right)}^{10}\cdot {\left(x+4\right)}^{10}\cdot {\left(x-2\right)}^{10}}=$

$=\underset{x\to 2}{\lim }\frac{{\left(x+1\right)}^{20}\cdot {\left(x-2\right)}^{20}}{{\left(x-2\right)}^{20}\cdot {\left(x+4\right)}^{10}}=$   

$=\underset{x\to 2}{\lim }\frac{{\left(x+1\right)}^{20}}{{\left(x+4\right)}^{10}}=$  

$=\frac{{\left(2+1\right)}^{20}}{{\left(2+4\right)}^{10}}=$

$=\frac{3^{20}}{6^{10}}=\frac{3^{10}\cdot 3^{10}}{2^{10}\cdot 3^{10}}=$   

$=\frac{3^{10}}{2^{10}}={\left(\frac{3}{2}\right)}^{10}$  

---

h)

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

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

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

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

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

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

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

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

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

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

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

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

$=\frac{-\left(3+2+1\right)}{3\cdot 2\cdot 2}=\frac{-6}{12}=-\frac{1}{2}$