Ответ:

1)

[tex]\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 3} \dfrac{x^{2} - 2x - 3}{x^{2} - 5x + 6} =4 } }[/tex]

2)

[tex]\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 3} \dfrac{x^{3} - 3x - 2}{x^{3} - 8} = \frac{16}{19} } }[/tex]

3)

[tex]\boxed{ \boldsymbol{ \displaystyle \lim_{x \to 2} \dfrac{3x^{4} - 5x^{2} - 12x - 4}{x^{2} - 4} = 16 } }[/tex]

4)

[tex]\boxed{ \boldsymbol{ \lim_{x \to-1} \dfrac{x^{5} + 1}{x + 1} =5 } }[/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 3} \dfrac{x^{2} - 2x - 3}{x^{2} - 5x + 6} = \bigg [\frac{0}{0} \bigg ] =\lim_{x \to 3} \dfrac{(x^{2} - 2x - 3)'}{(x^{2} - 5x + 6)'} = \lim_{x \to 3} \dfrac{2x - 2}{2x - 5} = \dfrac{2 \cdot 3 - 2}{2 \cdot 3 - 5} =[/tex]

[tex]= \dfrac{2 \cdot 3 - 2}{2 \cdot 3 - 5} = \dfrac{6 - 2}{6 - 5} = \dfrac{4}{1} = 4[/tex]

2)

[tex]\displaystyle \lim_{x \to 3} \dfrac{x^{3} - 3x - 2}{x^{3} - 8} = \dfrac{3^{3} - 3 \cdot 3 - 2}{3^{3} - 8} = \frac{27 - 9 - 2}{27 - 8} = \frac{16}{19}[/tex]

3)

[tex]\displaystyle \lim_{x \to 2} \dfrac{3x^{4} - 5x^{2} - 12x - 4}{x^{2} - 4} = \bigg [\frac{0}{0} \bigg ] = \lim_{x \to 2} \dfrac{(3x^{4} - 5x^{2} - 12x - 4)'}{(x^{2} - 4)'} =[/tex]

[tex]= \lim_{x \to 2} \dfrac{12x^{3} - 10x - 12}{2x} = \dfrac{12 \cdot 2^{3} - 10 \cdot 2 - 12}{2 \cdot 2} =\dfrac{12 \cdot 8 - 20 - 12}{4} =[/tex]

[tex]= 12 \cdot 2 - 5 - 3 = 24 - 8 =16[/tex]

4)

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