The correct solution to the limits of x in the tiles can be seen below.
- [tex]\mathbf{ \lim_{x \to 9^+} (\dfrac{|x-9|}{-x^2-34+387}) }[/tex][tex]\mathbf{ = -\dfrac{1}{52} }[/tex]
- [tex]\mathbf{ \lim_{x \to 8^-} (\dfrac{8-x}{|-x^2-63x+568|}) }[/tex][tex]\mathbf{=\dfrac{1}{79} }[/tex]
- [tex]\mathbf{ \lim_{x \to 7^+} (\dfrac{|-x^2-17x+168| }{x-7}) }[/tex]= -31
- [tex]\mathbf{ \lim_{x \to 6^-} (\dfrac{|x-6| }{-x^2-86x+552}) }[/tex][tex]\mathbf{ =\dfrac{1}{98}}[/tex]
What are the corresponding limits of x?
The limits of x approaching a given number of a quadratic equation can be determined by knowing the value of x at that given number and substituting the value of x into the quadratic equation.
From the given diagram, we have:
1.
[tex]\mathbf{ \lim_{x \to 9^+} (\dfrac{|x-9|}{-x^2-34+387}) }[/tex]
So, x - 9 is positive when x → 9⁺. Therefore, |x -9) = x - 9
[tex]\mathbf{ \lim_{x \to 9^+} (\dfrac{x-9}{-x^2-34+387}) }[/tex]
Simplifying the quadratic equation, we have:
[tex]\mathbf{ \lim_{x \to 9^+} (-\dfrac{1}{x+43}) }[/tex]
Replacing the value of x = 9
[tex]\mathbf{ = (-\dfrac{1}{9+43}) }[/tex]
[tex]\mathbf{ = -\dfrac{1}{52} }[/tex]
2.
[tex]\mathbf{ \lim_{x \to 8^-} (\dfrac{8-x}{|-x^2-63x+568|}) }[/tex]
- -x²-63x+568 is positive when x → 8⁻.
Thus |-x²-63x+568| = -x²-63x+568
[tex]\mathbf{ \lim_{x \to 8^-} (\dfrac{1}{x+71}) }[/tex]
[tex]\mathbf{=\dfrac{1}{8+71} }[/tex]
[tex]\mathbf{=\dfrac{1}{79} }[/tex]
3.
[tex]\mathbf{ \lim_{x \to 7^+} (\dfrac{|-x^2-17x+168| }{x-7}) }[/tex]
- x -7 is positive, therefore |x-7| = x - 7
[tex]\mathbf{ \lim_{x \to 7^+} (\dfrac{-x^2-17x+168 }{x-7}) }[/tex]
[tex]\mathbf{ \lim_{x \to 7^+} (-x-24)}[/tex]
[tex]\mathbf{ \lim_{x \to 7^+} (-7-24)}[/tex]
= -31
4.
[tex]\mathbf{ \lim_{x \to 6^-} (\dfrac{|x-6| }{-x^2-86x+552}) }[/tex]
- x-6 is negative when x → 6⁻. Therefore, |x-6| = -x + 6
[tex]\mathbf{ \lim_{x \to 6^-} (\dfrac{-x+6 }{-x^2-86x+552}) }[/tex]
[tex]\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{x+92}) }[/tex]
[tex]\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{6+92}) }[/tex]
[tex]\mathbf{ =\dfrac{1}{98}}[/tex]
Learn more about calculating the limits of x here:
https://brainly.com/question/1444047
#SPJ1
