Oblicz granice:- Zadanie 2.24: Matematyka 3. Zakres rozszerzony. Po gimnazjum

Rozwiązanie

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

$b) x \to + \infty lim \frac{6 x + 5}{- x ^{2} + 2 x + 1} = x \to + \infty lim \frac{x ^{2} ( \frac{6}{x} + \frac{5}{x ^{2}} )}{x ^{2} ( - 1 + \frac{2}{x} + \frac{1}{x ^{2}} )} = x \to + \infty lim \frac{\frac{6}{x} + \frac{5}{x ^{2}}}{- 1 + \frac{2}{x} + \frac{1}{x ^{2}}} = \frac{0}{- 1} = 0$

$c) x \to - \infty lim \frac{1 - 3 x + x ^{2}}{2 x ^{2} - 4 x - 9} = x \to - \infty lim \frac{x ^{2} ( \frac{1}{x ^{2}} - \frac{3}{x} + 1 )}{x ^{2} ( 2 - \frac{4}{x} - \frac{9}{x ^{2}} )} = x \to - \infty lim \frac{\frac{1}{x ^{2}} - \frac{3}{x} + 1}{2 - \frac{4}{x} - \frac{9}{x ^{2}}} = \frac{1}{2}$

$d) x \to - \infty lim \frac{8}{4 x ^{2} - 7 x + 15} = x \to - \infty lim \frac{x ^{2} ( \frac{8}{x ^{2}} )}{x ^{2} ( 4 - \frac{7}{x} + \frac{15}{x ^{2}} )} = x \to - \infty lim \frac{\frac{8}{x ^{2}}}{4 - \frac{7}{x} + \frac{15}{x ^{2}}} = \frac{0}{4} = 0$

$e) x \to + \infty lim \frac{( 2 x - 1 ) ^{2}}{( 1 - 3 x ) ^{2}} = x \to + \infty lim \frac{4 x ^{2} - 4 x + 1}{1 - 6 x + 9 x ^{2}} = x \to + \infty lim \frac{x ^{2} ( 4 - \frac{4}{x} + \frac{1}{x ^{2}} )}{x ^{2} ( \frac{1}{x ^{2}} - \frac{6}{x} + 9 )} =$

$= x \to + \infty lim \frac{4 - \frac{4}{x} + \frac{1}{x ^{2}}}{\frac{1}{x ^{2}} - \frac{6}{x} + 9} = \frac{4}{9}$

$f) x \to + \infty lim \frac{- ( 3 x + 1 ) ^{2}}{x ^{2} - ( 4 x + 2 ) ^{2}} = x \to + \infty lim \frac{- ( 9 x ^{2} + 6 x + 1 )}{x ^{2} - ( 16 x ^{2} + 16 x + 4 )} = x \to + \infty lim \frac{- 9 x ^{2} - 6 x - 1}{x ^{2} - 16 x ^{2} - 16 x - 4} =$

$= x \to + \infty lim \frac{- 9 x ^{2} - 6 x - 1}{- 15 x ^{2} - 16 x - 4} = x \to + \infty lim \frac{- x ^{2} ( 9 + \frac{6}{x} + \frac{1}{x ^{2}} )}{- x ^{2} ( 15 + \frac{16}{x} + \frac{4}{x ^{2}} )} = x \to + \infty lim \frac{9 + \frac{6}{x} + \frac{1}{x ^{2}}}{15 + \frac{16}{x} + \frac{4}{x ^{2}}} = \frac{9}{15} = \frac{3}{5}$

Magdalena Matusik

Nauczycielka matematyki

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