a)

  $f\left(x\right)=\frac{x^3-4}{2x-7}$   

 

$2x-7\ne 0$  

$2x\ne 7:2$  

$x\ne 3,5$   - a więc funkcja jest ciągła w  $x=2$       

  

$\underset{x\to 2}{\lim }\frac{x^3-4}{2x-7}=\frac{2^3-4}{2\cdot 2-7}=\frac{8-4}{4-7}=\frac{4}{-3}=-\frac{4}{3}$  

 

 

b)

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

 

 

c)  

$\underset{x\to 2}{\lim }\frac{8-x^3}{12-3x^2}=\underset{x\to 2}{\lim }\frac{2^3-x^3}{3\cdot \left(4-x^2\right)}=\underset{x\to 2}{\lim }\frac{\left(2-x\right)\left(2^2+2x+x^2\right)}{3\cdot \left(2-x\right)\left(2+x\right)}=\underset{x\to 2}{\lim }\frac{\left(4+2x+x^2\right)}{3\cdot \left(2+x\right)}=\frac{4+2\cdot 2+2^2}{3\cdot \left(2+2\right)}=\frac{4+4+4}{12}=\frac{12}{12}=1$  

 

d)

$\underset{x\to -2}{\lim }\frac{x^2-2x-8}{x^2-x-6}$  

 

$x^2-2x-8=0$  

${\Delta }_1={\left(-2\right)}^2-4\cdot 1\cdot \left(-8\right)=4+32=36$  

 

$x_1=\frac{2+6}{2}=\frac{8}{2}=4$  

$x_2=\frac{2-6}{2}=-\frac{4}{2}=-2$  

 

$x^2-x-6=0$  

${\Delta }_2={\left(-1\right)}^2-4\cdot 1\cdot \left(-6\right)=1+24=25$  

 

$x_1=\frac{1+5}{2}=\frac{6}{2}=3$  

$x_2=\frac{1-5}{2}=-\frac{4}{2}=-2$   

    

 

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

 

 

e)

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

 

$x^3-1=x^3-1^3=\left(x-1\right)\left(x^2+x+1\right)$  

 

$4x^2-7x+3=0$  

$\Delta ={\left(-7\right)}^2-4\cdot 4\cdot 3=49-48=1$  

 

$x_1=\frac{7-1}{4\cdot 2}=\frac{6}{8}=\frac{3}{4}$  

 

$x_2=\frac{7+1}{4\cdot 2}=\frac{8}{8}=1$   

       

 

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

 

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

 

 

f)

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

 

$9x^2-1=\left(3x-1\right)\left(3x+1\right)=3\cdot \left(x-\frac{1}{3}\right)\cdot 3\cdot \left(x+\frac{1}{3}\right)=9\cdot \left(x-\frac{1}{3}\right)\left(x+\frac{1}{3}\right)$  

 

$3x^2+2x-1=0$  

$\Delta =2^2-4\cdot 3\cdot \left(-1\right)=4+12=16$  

 

$x_1=\frac{-2-4}{2\cdot 3}=\frac{-6}{6}=-1$  

 

$x_2=\frac{-2+4}{2\cdot 3}=\frac{2}{6}=\frac{1}{3}$  

        

 

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

 

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

 

 

f)

$\underset{x\to 5}{\lim }\frac{25x-x^3}{x^2-4x-5}$  

 

$25x-x^3=x\left(25-x^2\right)=x\left(5-x\right)\left(5+x\right)$  

 

$x^2-4x-5=0$  

$\Delta ={\left(-4\right)}^2-4\cdot 1\cdot \left(-5\right)=16+20=36$  

 

$x_1=\frac{4+6}{2\cdot 1}=\frac{10}{2}=5$  

 

$x_2=\frac{4-6}{2\cdot 1}=\frac{-2}{2}=-1$   

         

 

$\underset{x\to 5}{\lim }\frac{25x-x^3}{x^2-4x-5}=\underset{x\to 5}{\lim }\frac{x\left(5-x\right)\left(5+x\right)}{\left(x-5\right)\left(x+1\right)}=\underset{x\to 5}{\lim }\frac{-x\left(x-5\right)\left(5+x\right)}{\left(x-5\right)\left(x+1\right)}=\underset{x\to 5}{\lim }\frac{-x\left(5+x\right)}{x+1}=$  

 

$=-\frac{5\cdot \left(5+5\right)}{5+1}=-\frac{5\cdot 10}{6}=-\frac{50}{6}=-\frac{25}{3}$     

 

 

h)

$\underset{x\to -4}{\lim }\frac{x^3+3x^2-4x}{x^2+7x+12}$  

 

$x^3+3x^2-4x=0$  

$x\left(x^2+3x-4\right)=0$  

 

${\Delta }_1=3^2-4\cdot 1\cdot \left(-4\right)=9+16=25$  

$x_1=\frac{-3+5}{2\cdot 1}=\frac{2}{2}=1$  

$x_2=\frac{-3-5}{2\cdot 1}=-\frac{8}{2}=-4$  

 

 

$x^2+7x+12=0$  

${\Delta }_2=7^2-4\cdot 1\cdot 12=49-48=1$  

 

$x_1=\frac{-7+1}{2}=-\frac{6}{2}=-3$  

 

$x_2=\frac{-7-1}{2}=-\frac{8}{2}=-4$  

 

 

$\underset{x\to -4}{\lim }\frac{x^3+3x^2-4x}{x^2+7x+12}=\underset{x\to -4}{\lim }\frac{x\left(x-1\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)}=\underset{x\to -4}{\lim }\frac{x\left(x-1\right)}{x+3}=$  

 

$=\frac{-4\cdot \left(-4-1\right)}{-4+3}=\frac{-4\cdot \left(-5\right)}{-1}=-20$  

 

 

i)

$\underset{x\to \frac{1}{2}}{\lim }\frac{16x^3-2}{8x^2-6x+1}$  

 

$16x^3-2=2\left(8x^3-1\right)=2\left({\left(2x\right)}^3-1^3\right)=2\left(2x-1\right)\left({\left(2x\right)}^2+2x+1^2\right)=2\left(2x-1\right)\left(4x^2+2x+1\right)=2\cdot 2\cdot \left(x-\frac{1}{2}\right)\left(4x^2+2x+1\right)$   

 

 

$8x^2-6x+1=0$  

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

 

$x_1=\frac{6+2}{2\cdot 8}=\frac{8}{16}=\frac{1}{2}$  

 

$x_2=\frac{6-2}{2\cdot 8}=\frac{4}{16}=\frac{1}{4}$   

        

    

$\underset{x\to \frac{1}{2}}{\lim }\frac{16x^3-2}{8x^2-6x+1}=\underset{x\to \frac{1}{2}}{\lim }\frac{4\left(x-\frac{1}{2}\right)\left(4x^2+2x+1\right)}{8\left(x-\frac{1}{2}\right)\left(x-\frac{1}{4}\right)}=\underset{x\to \frac{1}{2}}{\lim }\frac{4x^2+2x+1}{2\left(x-\frac{1}{4}\right)}=$   

 

$=\frac{4\cdot \frac{1}{2^2}+2\cdot \frac{1}{2}+1}{2\cdot \left(\frac{1}{2}-\frac{1}{4}\right)}=\frac{1+1+1}{1-\frac{1}{2}}=\frac{3}{\frac{1}{2}}=6$