Answer:
There is a vertical Asymptote at x = 5 because [tex]\lim_{x \to 5^{-} } G(x) = \infty\\lim_{x \to 5^{+} } G(x) = -\infty[/tex]
There is a vertical Asymptote at x = -5 because [tex]\lim_{x \to -5^{-} } G(x) = \infty\\\lim_{x \to -5^{+} } G(x) = -\infty[/tex]
Step-by-step explanation:
The exact question is as follows :
Given - G(x) = -7(x-5)^2(x+6)/(x-5)(x+5)
To find - Which statement make correctly uses limits determine a vertical Asymptote of G(x)
Solution -
Vertical Asymptotes are vertical lines which correspond to the zeroes of the denominator of a rational function.
Here the given function is the rational function
Denominator of G(x) = (x - 5)(x + 5)
So,
Put denominator = 0, we get
(x - 5)(x + 5) = 0
⇒x = 5, -5
∴ we get
G(x) has vertical Asymptotes at x = 5 and x = -5
Now,
At x= 5
[tex]\lim_{x \to 5^{-} } G(x) = \infty\\\lim_{x \to 5^{+} } G(x) = -\infty[/tex]
And
At x = -5
[tex]\lim_{x \to -5^{-} } G(x) = \infty\\\lim_{x \to -5^{+} } G(x) = -\infty[/tex]
∴ we get
There is a vertical Asymptote at x = 5 because [tex]\lim_{x \to 5^{-} } G(x) = \infty\\lim_{x \to 5^{+} } G(x) = -\infty[/tex]
There is a vertical Asymptote at x = -5 because [tex]\lim_{x \to -5^{-} } G(x) = \infty\\\lim_{x \to -5^{+} } G(x) = -\infty[/tex]
