Verified answer

Ответ:

1)

[tex]\boxed{ \boldsymbol{ \lim_{x \to 1} (2x^{2} - 3x - 1) = -2} }[/tex]

2)

[tex]\boxed{ \boldsymbol{ \lim_{x \to -1} x^{4} = 1}}[/tex]

3)

[tex]\boxed{ \boldsymbol{ \lim_{x \to 2} (x^{3} - 3x - 2) = 0}}[/tex]

4)

[tex]\boxed{ \boldsymbol{ \lim_{x \to 3} \dfrac{2x - 1}{3x - 2} =\dfrac{5}{7}} }[/tex]

5)

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

6)

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

Примечание:

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

Объяснение:

1) [tex]\lim_{x \to 1} (2x^{2} - 3x - 1) = 2\cdot 1^{2} - 3 \cdot 1 - 1 = 2 - 3 -1 = 2 - 4 = -2[/tex]

2) [tex]\lim_{x \to -1} x^{4} = (-1)^{4} = 1[/tex]

3) [tex]\lim_{x \to 2} (x^{3} - 3x - 2) = 2^{3} - 3 \cdot 2 - 2 = 8 - 6 - 2 = 2 - 2 = 0[/tex]

4) [tex]\lim_{x \to 3} \dfrac{2x - 1}{3x - 2} = \dfrac{2 \cdot 3 - 1}{3 \cdot 3 - 2}=\dfrac{6 - 1}{9 - 2} = \dfrac{5}{7}[/tex]

5)  [tex]\lim_{x \to 0} \dfrac{x^{2} - 3x + 5}{x^{2} + 2x - 1} =\dfrac{0^{2} - 3 \cdot 0 + 5}{0^{2} + 2 \cdot 0 - 1} = \dfrac{5}{-1} = -5[/tex]

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

[tex]= 0,5 - 3 = -2,5[/tex]