Ответ:

1)

[tex]\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 1} \dfrac{x^{2} - 3x + 2}{x^{2} - 4x + 3} = 0,5 } }[/tex]

2)

[tex]\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 1} \dfrac{x^{4} - 3x + 2}{x^{5} - 4x + 3} =1 } }[/tex]

3)

[tex]\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 0} \dfrac{(1 + x)^{4} - (1 + 4x)}{x^{2} + x^{4}} = 6 } }[/tex]

4)

[tex]\boxed{ \boldsymbol{ \lim_{x \to 1} \dfrac{x^{5} - 1}{x^{3} - 1} = \frac{5}{3} } }[/tex]

Примечание:

[tex]\lim_{x \to a} f(a) = f(a)[/tex] если  [tex]f[/tex] существует в точке [tex]a[/tex]

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Если:

1) [tex]\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0[/tex]

2) функции [tex]f(x)[/tex] и [tex]g(x)[/tex] дифференцируемы в окрестности точки [tex]x = a[/tex]

3) [tex]\exists \lim_{x \to a} \dfrac{f'(x)}{g'(x)}[/tex]

По правилу Лопиталя:

[tex]\boxed{ \lim_{x \to a} \dfrac{f(x)}{g(x)} = \bigg [ \dfrac{0}{0} \bigg ] = \lim_{x \to a} \dfrac{f'(x)}{g'(x)} }[/tex]

Объяснение:

1)

[tex]\displaystyle \lim_{x \to 1} \dfrac{x^{2} - 3x + 2}{x^{2} - 4x + 3} = \bigg [\frac{0}{0} \bigg ] = \lim_{x \to 1} \dfrac{(x^{2} - 3x + 2)'}{(x^{2} - 4x + 3)'} = \lim_{x \to 1} \dfrac{2x - 3}{2x - 4} = \dfrac{2 \cdot 1 - 3}{2 \cdot 1 - 4}=[/tex]

[tex]= \dfrac{2 - 3}{2 - 4}= \dfrac{-1}{-2} = 0,5[/tex]

2)

[tex]\displaystyle \lim_{x \to 1} \dfrac{x^{4} - 3x + 2}{x^{5} - 4x + 3} = \bigg [\frac{0}{0} \bigg ] =\lim_{x \to 1} \dfrac{(x^{4} - 3x + 2)'}{(x^{5} - 4x + 3)'} = \lim_{x \to 1} \dfrac{4x^{3} - 3}{5x^{4} - 4} = \dfrac{4 \cdot 1^{3} - 3}{5 \cdot 1^{4} - 4} =[/tex]

[tex]= \dfrac{4 - 3}{5 - 4} = \dfrac{1}{1} = 1[/tex]

3)

[tex]\displaystyle \lim_{x \to 0} \dfrac{(1 + x)^{4} - (1 + 4x)}{x^{2} + x^{4}} = \displaystyle \lim_{x \to 0} \dfrac{(1 + x)^{4} - 4x - 1}{x^{2} + x^{4}} = \bigg [\dfrac{0}{0} \bigg ] = \lim_{x \to 0} \dfrac{((1 + x)^{4} - 4x - 1)'}{(x^{2} + x^{4})'} =[/tex]

[tex]= \lim_{x \to 0} \dfrac{4(1 + x)^{3}(x + 1)' - 4}{2x + 4x^{3}} = \lim_{x \to 0} \dfrac{4(1 + x)^{3} - 4}{2x + 4x^{3}} = \bigg [\dfrac{0}{0} \bigg ] =[/tex]

[tex]= \lim_{x \to 0} \dfrac{(4(1 + x)^{3} - 4)'}{(2x + 4x^{3})'} = \lim_{x \to 0} \dfrac{12(1 + x)^{2} }{2 + 12x^{2}} = \dfrac{12(1 + 0)^{2} }{2 + 12 \cdot 0^{2}} = \dfrac{12}{2} = 6[/tex]

4)

[tex]\lim_{x \to 1} \dfrac{x^{5} - 1}{x^{3} - 1} = \bigg [\dfrac{0}{0} \bigg] = \lim_{x \to 1} \dfrac{(x^{5} - 1)'}{(x^{3} - 1)'} = \lim_{x \to 1} \dfrac{5x^{4}}{3x^{2} } = \lim_{x \to 1} \dfrac{5x^{2}}{3 } =[/tex]

[tex]= \dfrac{5 \cdot 1^{2}}{3 } = \dfrac{5}{3}[/tex]