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Ответ:

Пределы:

1) [tex]\boxed{ \boldsymbol { \displaystyle \lim_{x \to 1} \bigg(\frac{1}{1 - x} - \frac{3}{1 - x^{3}} \bigg) = -1 } }[/tex]

2) [tex]\boxed{ \boldsymbol { \displaystyle \lim_{x \to 2} \bigg(\frac{2}{x^{2} - 2x} - \frac{3}{x^{2} - 3x +2} \bigg) = 0,5 } }[/tex]

Объяснение:

38.5

1)

[tex]\displaystyle \lim_{x \to 1} \bigg(\frac{1}{1 - x} - \frac{3}{1 - x^{3}} \bigg) = -1[/tex]

а) Преобразуем выражение [tex]\displaystyle \bigg(\frac{1}{1 - x} - \frac{3}{1 - x^{3}} \bigg) :[/tex]

[tex]\displaystyle \bigg(\frac{1}{1 - x} - \frac{3}{1 - x^{3}} \bigg) = \frac{1}{1 - x} - \frac{3}{(1 - x)(1 + x + x^{2})} = \frac{1 + x + x^{2}-3}{(1 - x)(1 + x + x^{2})} =[/tex][tex]\displaystyle = \frac{x + x^{2} - 2}{(1 - x)(1 + x + x^{2})}[/tex]

б) разложим на множители выражение [tex]x + x^{2} - 2:[/tex]

[tex]x + x^{2} - 2 = 0[/tex]

[tex]D = 1 - 4 \cdot 1 \cdot (-2) =1 + 8 = 9 = 3^{2}[/tex]

[tex]\displaystyle x_{1} = \frac{-1 + 3}{2} = \frac{2}{2} = 1[/tex]

[tex]\displaystyle x_{2} = \frac{-1 - 3}{2} = \frac{-4}{2} = -2[/tex]

[tex]x + x^{2} - 2 = 1(x + 2)(x - 1) = (x - 1)(x + 2)[/tex]

в) воспользуемся результатами пункта а) и б)

[tex]\displaystyle \frac{x + x^{2} - 2}{(1 - x)(1 + x + x^{2})} = \frac{(x - 1)(x + 2)}{(1 - x)(1 + x + x^{2})} = \frac{-(1 - x)(x + 2)}{(1 - x)(1 + x + x^{2})} =[/tex]

[tex]\displaystyle = \frac{-(x + 2)}{(1 + x + x^{2})}[/tex]

[tex]\displaystyle \lim_{x \to 1} \bigg(\frac{1}{1 - x} - \frac{3}{1 - x^{3}} \bigg) = \lim_{x \to 1} \frac{-(x + 2)}{(1 + x + x^{2})} = \frac{-(1 + 2)}{(1 + 1 + 1^{2})} = \frac{-3}{3} =-1[/tex]

2)

[tex]\displaystyle \lim_{x \to 2} \bigg(\frac{2}{x^{2} - 2x} - \frac{3}{x^{2} - 3x +2} \bigg) = 0,5[/tex]

а) Преобразуем выражение [tex]\displaystyle \bigg(\frac{2}{x^{2} - 2x} - \frac{3}{x^{2} - 3x +2} \bigg):[/tex]

[tex]\displaystyle \bigg(\frac{2}{x^{2} - 2x} - \frac{3}{x^{2} - 3x +2} \bigg) = \frac{2}{x(x - 2)} - \frac{3}{x^{2} - 3x +2}[/tex]

б) разложим на множители выражение [tex]x^{2} - 3x + 2:[/tex]

[tex]x^{2} - 3x + 2 = 0[/tex]

[tex]D = 9 - 4 \cdot 1 \cdot 2 = 9 - 8 = 1 = 1^{2}[/tex]

[tex]\displaystyle x_{1} = \frac{3 + 1}{2} = \frac{4}{2} = 2[/tex]

[tex]\displaystyle x_{2} = \frac{3 - 1}{2} = \frac{2}{2} = 1[/tex]

[tex]x^{2} - 3x + 2 = 1(x - 1)(x - 2) = (x - 1)(x - 2)[/tex]

в) воспользуемся результатами пункта а) и б)

[tex]\displaystyle \frac{2}{x(x - 2)} - \frac{3}{x^{2} - 3x +2} = \frac{2}{x(x - 2)} - \frac{3}{(x - 1)(x - 2)} = \frac{2(x - 1) - 3x}{x(x - 1)(x - 2)} =[/tex]

[tex]\displaystyle = \frac{2x - 2 - 3x}{x(x - 1)(x - 2)} = \frac{x - 2}{x(x - 1)(x - 2)} = \frac{1}{x(x - 1)}[/tex]

[tex]\displaystyle \lim_{x \to 2} \bigg(\frac{2}{x^{2} - 2x} - \frac{3}{x^{2} - 3x +2} \bigg) =\lim_{x \to 2} \frac{1}{x(x - 1)} = \frac{1}{2(2 -1)} = \frac{1}{2} = 0,5[/tex]