Oblicz granicę. zadanie w załączniku.... Question from @Qetuop17

January 2019 1 22 Report

Oblicz granicę. zadanie w załączniku

Mathēmatikós 7.71

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

$\lim_{x \to 1} ( x^{6} + 2x^{5} + 3x^{4} + 4x^{3}+5x^{2}+6x+7 ) = \\ \\ =1^{6}+2*1^{5} +3*1^{4} + 4*1^{3}+5*1^{2}+6*1+7 = 28$

7.72

Na początku upraszczamy ułamki.

$\frac{3}{1-x^{3}} - \frac{4}{1-x^{4}} = \frac{3(1-x^{4})}{(1-x^{3})(1-x^{4})} - \frac{4(1-x^{3})}{(1-x^{3})(1-x^{4})} = \\ \\ \frac{3(1-x^{2})(1+x^{2})}{(1-x)(1+x+x^{2})(1-x^{2})(1+x^{2})} - \frac{4(1-x)(1+x+x^{2})}{(1-x)(1+x+x^{2})(1-x^{2})(1+x^{2})} =$

$=\frac{3(1-x)(1+x)(1+x^{2})}{(1-x)(1+x+x^{2})(1-x)(1+x)(1+x^{2})} - \frac{4(1-x)(1+x+x^{2})}{(1-x)(1+x+x^{2})(1-x)(1+x)(1+x^{2})} = \\ \\ = \frac{(1-x)[3(1+x)(1+x^{2}] - (1-x)[4(1+x+x^{2}]}{(1-x)^{2}(1+x+x^{2})(1+x)(1+x^{2})} =$

$= \frac{(1-x)[3(1+x+x^{2}+x^{3}) - 4(1+x+x^{2})]}{(1-x)^{2}(1+x+x^{2})(1+x)(1+x^{2})} = \frac{3 + 3x + 3x^{2} + 3x^{3} - 4 - 4x - 4x^{2}}{(1-x)(1+x+x^{2})(1+x)(1+x^{2})}= \\ \\ = \frac{3x^{3} - x^{2} - x - 1}{(1-x)(1+x+x^{2})(1+x)(1+x^{2})} = \frac{3x^{3} - 3x^{2} + 2x^{2} - 2x + x - 1}{(1-x)(1+x+x^{2})(1+x)(1+x^{2})} =$

$= \frac{3x^{2}(x-1) + 2x(x-1) + (x-1)}{(1-x)(1+x+x^{2})(1+x)(1+x^{2})} = \frac{(x-1)(3x^{2}+2x+1)}{(1-x)(1+x+x^{2})(1+x)(1+x^{2})} = \\ \\ = \frac{3x^{2}+2x+1}{(x^{2}+x+1)(x+1)(x^{2}+1)}$

Zatem

$\lim_{x \to 1} ( \frac{3}{1-x^{3}} - \frac{4}{1-x^{4}} ) = \lim_{x \to 1} \frac{3x^{2}+2x+1}{(x^{2}+x+1)(x+1)(x^{2}+1)} = \\ \\ = \frac{3*1^{2}+2*1+1}{(1^{2}+1+1)(1+1)(1^{2}+1)} = \frac{3+2+1}{3*2*2} = \frac{6}{12} = \frac{1}{2}$

7.73

Ponownie uprościmy ułamek.

$\frac{x^{6} - 6x+5}{(x-1)^{2}} = \frac{x^{6}- x^{5} + x^{5} - x^{4} + x^{4} - x^{3} + x^{3} -x^{2} + x^{2} - x - 5x + 5 }{(x-1)^{2}} = \\ \\ = \frac{x^{5}(x-1) + x^{4}(x-1) + x^{3}(x-1) + x^{2}(x-1) + x(x-1) - 5(x-1)}{(x-1)^{2}} = \\ \\ = \frac{(x-1)(x^{5}+x^{4}+x^{3}+x^{2}+x-5)}{(x-1)^{2}} = \frac{x^{5} - x^{4} + 2x^{4} - 2x^{3} + 3x^{3} - 3x^{2} + 4x^{2} - 4x + 5x - 5}{(x-1)} =$

$=\frac{x^{4}(x-1) + 2x^{3}(x-1) + 3x^{2}(x-1) + 4x(x-1) + 5(x-1)}{(x-1)} = \\ \\ = \frac{(x-1)(x^{4}+2x^{3} + 3x^{2} + 4x+5)}{x-1} = x^{4} + 2x^{3} + 3x^{2} + 4x+5$

Zatem

$\lim_{x \to 1} \frac{x^{6} - 6x+5}{(x-1)^{2}} = \lim_{x \to 1} x^{4} + 2x^{3} + 3x^{2} + 4x+1 = \\ \\ 1^{4} + 2*1^{3} + 3*1^{2} + 4*1 +5 = 15$

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