Rozwiązanie:
[tex]a)[/tex]
[tex]$ \lim_{x \to 0} \frac{x^{3}-5x^{2}}{x^{2}} = \lim_{x \to 0} (x-5)=-5[/tex]
[tex]b)[/tex]
[tex]$ \lim_{x \to 3} \frac{x^{2}-9}{x^{3}-27} =\lim_{x \to 3} \frac{(x-3)(x+3)}{(x-3)(x^{2}+3x+9)} =\lim_{x \to 3} \frac{x+3}{x^{2}+3x+9} =\frac{3+3}{9+9+9} =\frac{2}{9}[/tex]
[tex]c)[/tex]
[tex]$\lim_{x \to 1} \frac{4x^{2}-4}{x^{3}+3x^{2}-7x+3} =\lim_{x \to 1} \frac{4(x-1)(x+1)}{(x-1)(x^{2}+4x-3)} =\lim_{x \to 1} \frac{4(x+1)}{x^{2}+4x-3} =\frac{8}{2} =4[/tex]
