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PLEASE HELP!!!!!!!!!!!!!!!!!!!!!

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!-example-1
User Marguerita
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1 Answer

5 votes

Explanation:

the last 2 answer options are completely useless and outright nonsense, because all limits are going against +4.

so, an asymptote at x = -4 is not an option anywhere.

g(x) is always y, and the limit of the function is the limit of the y values.

we just evaluate for what x value is then y reaching that limit. but the sign of the y limit has no direct impact on the sign of the x value.

it is true that the case x = -4 needs to be checked too, but it turns out this just delivers g(-4) = 0.

now, about x = 4 :

when coming from the left side of +4 (increasing x), what is the sign of the function ?

we can simply use x = 3 to test this :

g(3) = 2(3+4)²/(3² - 16) = 2×49/-7

so, it is negative.

that means when getting closer and closer to +4 coming from that side we what get negative values, and the limit is -infinity.

when coming from the right side of +4 (decreasing x), what is the sign of the function ?

we can simply use x = 5 to test this :

g(3) = 2(5+4)²/(5² - 16) = 2×81/9

so, it is positive.

that means when getting closer and closer to +4 coming from that side we what get positive values, and the limit is +infinity.

so, the second answer option is correct.

because of these y limits, there is a vertical asymptote at x = 4 that kind of "connects" -infinity and +infinity in y direction at that location and symbolizes the limits of y.

User Weibenfalk
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