Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]\lim_{x \to3}\frac{x^2-9}{3-x} =[\frac{0}{0} ]= \lim_{x \to3}\frac{(x-3)(x+3)}{-(x-3)} = \lim_{x \to 3} (-x-3)=-3-3=-6[/tex]
[tex]\lim_{x \to 5} \frac{125-x^3}{4x^2-100} =[\frac{0}{0} ]= \lim_{x \to 5} \frac{(5-x)(25+5x+x^2)}{4(x^-5)(x+5)} = \lim_{x \to5}\frac{-(x-5)(25+5x+x^2)}{4(x-5)(x+5)} = \lim_{x \to 5} \frac{-x^2-5x-25}{4x+20} =\frac{-25-25-25}{20+20} =-\frac{75}{40}=-1\frac{7}{8}[/tex]
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Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]\lim_{x \to3}\frac{x^2-9}{3-x} =[\frac{0}{0} ]= \lim_{x \to3}\frac{(x-3)(x+3)}{-(x-3)} = \lim_{x \to 3} (-x-3)=-3-3=-6[/tex]
[tex]\lim_{x \to 5} \frac{125-x^3}{4x^2-100} =[\frac{0}{0} ]= \lim_{x \to 5} \frac{(5-x)(25+5x+x^2)}{4(x^-5)(x+5)} = \lim_{x \to5}\frac{-(x-5)(25+5x+x^2)}{4(x-5)(x+5)} = \lim_{x \to 5} \frac{-x^2-5x-25}{4x+20} =\frac{-25-25-25}{20+20} =-\frac{75}{40}=-1\frac{7}{8}[/tex]