Функције – граничне вредности функција 5

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Одредити граничне вредности:

пр.14)   $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 6x - 7}}{{{x^2} - 5x + 4}}$

пр.15)   $\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 2{x^2} + 5x - 10}}{{{x^3} - 8{x^2} + 12x}}$

пр.16)   $\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x + 5} - \sqrt {2x + 2} }}{{{x^3} - 27}}$

пр.17)   $\mathop {\lim }\limits_{x \to 8} \frac{{\sqrt {x + 8} - 4}}{{2 - \sqrt[3]{x}}}$

пр.14)   

$$\begin{aligned}&\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 6x - 7}}{{{x^2} - 5x + 4}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + 7} \right)\left( {x - 1} \right)}}{{\left( {x - 4} \right)\left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + 7} \right)}}{{\left( {x - 4} \right)}} = \frac{{1 + 7}}{{1 - 4}} = \frac{8}{{ - 3}} \\ & \\ &\begin{array}{cccccccccccccccccccc} {{x^2} - 6x - 7 = 0}&{}&{}&{{x^2} - 5x + 4 = 0} \\ {{x_{1/2}} = \frac{{ - 6 \pm \sqrt {36 + 4 \cdot 7} }}{2}}&{}&{}&{{x_{1/2}} = \frac{{5 \pm \sqrt {25 - 4 \cdot 4} }}{2}} \\ {{x_{1/2}} = \frac{{ - 6 \pm \sqrt {36 + 28} }}{2}}&{}&{}&{{x_{1/2}} = \frac{{5 \pm \sqrt {25 - 16} }}{2}} \\ {{x_{1/2}} = \frac{{ - 6 \pm \sqrt {64} }}{2}}&{}&{}&{{x_{1/2}} = \frac{{5 \pm \sqrt 9 }}{2}} \\ {{x_{1/2}} = \frac{{ - 6 \pm 8}}{2}}&{}&{}&{{x_{1/2}} = \frac{{5 \pm 3}}{2}} \\ {{x_1} = - 7;{x_2} = 1}&{}&{}&{{x_1} = 4;{x_2} = 1} \\ {\left( {x + 7} \right)\left( {x - 1} \right) = 0}&{}&{}&{\left( {x - 4} \right)\left( {x - 1} \right) = 0} \end{array}\end{aligned}$$

пр.15)  

$$\begin{aligned}&\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 2{x^2} + 5x - 10}}{{{x^3} - 8{x^2} + 12x}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {{x^2} + 5} \right)}}{{x\left( {x - 2} \right)\left( {x - 6} \right)}} = \frac{{{2^2} + 5}}{{2\left( {2 - 6} \right)}} = \frac{9}{{ - 8}} \\ & \\ &{x^3} - 2{x^2} + 5x - 10 = {x^2}\left( {x - 2} \right) + 5\left( {x - 2} \right) = \left( {x - 2} \right)\left( {{x^2} + 5} \right) \\ &{x^3} - 8{x^2} + 12x = x\left( {{x^2} - 8x + 12} \right) = x\left( {x - 2} \right)\left( {x - 6} \right) \\ &{x^2} - 8x + 12 = 0 \\ &{x_{1/2}} = \frac{{8 \pm \sqrt {64 - 4 \cdot 12} }}{2} \\ &{x_{1/2}} = \frac{{8 \pm 4}}{2} \\ &{x_1} = 2;{x_2} = 6\end{aligned}$$

пр.16)  

$$\begin{aligned}&\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x + 5} - \sqrt {2x + 2} }}{{{x^3} - 27}} = \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x + 5} - \sqrt {2x + 2} }}{{{x^3} - 27}} \cdot \frac{{\sqrt {x + 5} + \sqrt {2x + 2} }}{{\sqrt {x + 5} + \sqrt {2x + 2} }} = \\ &= \mathop {\lim }\limits_{x \to 3} \frac{{x + 5 - 2x - 2}}{{\left( {{x^3} - 27} \right)\left( {\sqrt {x + 5} + \sqrt {2x + 2} } \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{{ - x + 3}}{{\left( {{x^3} - 27} \right)\left( {\sqrt {x + 5} + \sqrt {2x + 2} } \right)}} = \\ &= \mathop {\lim }\limits_{x \to 3} \frac{{x + 5 - 2x - 2}}{{\left( {{x^3} - 27} \right)\left( {\sqrt {x + 5} + \sqrt {2x + 2} } \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{{ - x + 3}}{{\left( {{x^3} - 27} \right)\left( {\sqrt {x + 5} + \sqrt {2x + 2} } \right)}} = \\ &= \mathop {\lim }\limits_{x \to 3} \frac{{ - \left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)\left( {\sqrt {x + 5} + \sqrt {2x + 2} } \right)}} = \\ &= \frac{{ - 1}}{{\left( {9 + 9 + 9} \right) \cdot \left( {\sqrt {3 + 5} + \sqrt {2 \cdot 3 + 2} } \right)}} = \frac{{ - 1}}{{27 \cdot \left( {2\sqrt 8 } \right)}} = \\ &= \frac{{ - 1}}{{27 \cdot 2 \cdot 2\sqrt 2 }} = \frac{{ - 1}}{{108\sqrt 2 }} = \frac{{ - 1}}{{108\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} = - \frac{{\sqrt 2 }}{{216}}\end{aligned}$$

пр.17)  

$$\begin{aligned}&\mathop {\lim }\limits_{x \to 8} \frac{{\sqrt {x + 8} - 4}}{{2 - \sqrt[3]{x}}} = \mathop {\lim }\limits_{x \to 8} \frac{{\sqrt {x + 8} - 4}}{{2 - \sqrt[3]{x}}} \cdot \frac{{\sqrt {x + 8} + 4}}{{\sqrt {x + 8} + 4}} \cdot \frac{{{2^2} + 2\sqrt[3]{x} + \sqrt[3]{{{x^2}}}}}{{{2^2} + 2\sqrt[3]{x} + \sqrt[3]{{{x^2}}}}} = \\ &= \mathop {\lim }\limits_{x \to 8} \frac{{\left( {x + 8 - 16} \right)\left( {4 + 2\sqrt[3]{x} + \sqrt[3]{{{x^2}}}} \right)}}{{\left( {{2^3} - x} \right)\left( {\sqrt {x + 8} + 4} \right)}} = \mathop {\lim }\limits_{x \to 8} \frac{{\left( {x - 8} \right)\left( {4 + 2\sqrt[3]{x} + \sqrt[3]{{{x^2}}}} \right)}}{{\left( {8 - x} \right)\left( {\sqrt {x + 8} + 4} \right)}} = \\ &= \mathop {\lim }\limits_{x \to 8} \frac{{\left( {4 + 2\sqrt[3]{x} + \sqrt[3]{{{x^2}}}} \right)}}{{\left( {\sqrt {x + 8} + 4} \right)}} = \frac{{4 + 4 + \sqrt[3]{{64}}}}{{ - \left( {\sqrt {16} + 4} \right)}} = \frac{{4 + 4 + 4}}{{ - \left( {4 + 4} \right)}} = - \frac{{12}}{8} = - 1\frac{1}{2}\end{aligned}$$