Question

Niste ajutor va rog!​

Answer 1

Explicație pas cu pas:

1.

$= {3}^{2} - 2 \cdot 3 + 5 = 9 - 6 + 5 = 8$

2.

$= 5 \cdot (- \infty) = - \infty$

3.

$= \dfrac{ - {1}^{2} + 3}{ - 2 \cdot 1 - 7} = - \dfrac{2}{9}$

4.

$\lim_{x \rightarrow - 4} \dfrac{3x + 12}{2 {x}^{2} - 32} = \lim_{x \rightarrow - 4} \dfrac{3(x + 4)}{2 ({x}^{2} - 16)} = \lim_{x \rightarrow - 4} \dfrac{3(x + 4)}{2 (x - 4)(x + 4)} = \lim_{x \rightarrow - 4} \dfrac{3}{2 (x - 4)} = \dfrac{3}{2 ( - 4 - 4)} = - \dfrac{3}{16}$

5.

$\lim_{x \rightarrow - \infty } \dfrac{ - 2 {x}^{4} + 9}{{x}^{4}} = \lim_{x \rightarrow - \infty }\dfrac{ - 2 {x}^{4}}{{x}^{4}} + \lim_{x \rightarrow - \infty } \dfrac{9}{{x}^{4}} = - 2 + 0 = - 2$

6.

$\lim_{x \rightarrow 2} \dfrac{ \sin(x - 2) }{{x}^{2} - 4} = \lim_{x \rightarrow 2} \dfrac{ \sin(x - 2) }{(x - 2)(x + 2)} = \lim_{x \rightarrow 2} \dfrac{ \sin(x - 2) }{x - 2} \cdot \lim_{x \rightarrow 2}\dfrac{1}{x + 2} = 1 \cdot \dfrac{1}{2 + 2} = \dfrac{1}{4}$