Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]\lim_{x \to 3} \frac{6x-18}{x-3} = \lim_{x \to 3} \frac{6(x-3)}{x-3} =6[/tex]
[tex]\lim_{x \to8} \frac{3x^2+2x-5}{6x^2-3x+4} =\frac{203}{364} =\frac{29}{52}[/tex]
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Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]\lim_{x \to 3} \frac{6x-18}{x-3} = \lim_{x \to 3} \frac{6(x-3)}{x-3} =6[/tex]
[tex]\lim_{x \to8} \frac{3x^2+2x-5}{6x^2-3x+4} =\frac{203}{364} =\frac{29}{52}[/tex]