Answer:
The correct option is;
b. y = x², [tex]y = \left | x \right |[/tex]
Step-by-step explanation:
The equations are
1) y = x
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )=-\infty[/tex]
2) y = x²
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )=+\infty[/tex]
3) y = x³
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )=-\infty[/tex]
4) [tex]y = \left | x \right |[/tex]
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )=+\infty[/tex]
5) y = 1/x
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )=0[/tex]
6) y = eˣ
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )\rightarrow 0[/tex]
7) y = √x
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )= \infty\sqrt{(-1)}[/tex]
8) y = ㏑x
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )= \infty\sqrt{(-1)}[/tex]
9) y = sin x
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )=[/tex] undefined
10) y = cos x
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )=[/tex] undefined
11) y int (x)
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )=-\infty[/tex]
12) [tex]y = \dfrac{1}{1 + e^{-x}}[/tex]
[tex]\lim_{x\rightarrow -\infty }f\left (x \right )=0[/tex]
Therefore, the correct options are y = x² and [tex]y = \left | x \right |[/tex]
