Explain why please! My AP test is Wednesday.
The function f is given by f(x) = sin ( (x + 1) / (x^2) ). Which of the following statements are true?
I. The graph of f has a horizontal asymptote at y = 0.
II. The graph of f has a horizontal asymptote at y = 1.
III. The graph of f has a vertical asymptote at x = 0.
The function f is given by f(x) = sin ( (x + 1) / (x^2) ). Which of the following statements are true?
I. The graph of f has a horizontal asymptote at y = 0.
II. The graph of f has a horizontal asymptote at y = 1.
III. The graph of f has a vertical asymptote at x = 0.
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Let's look at what happens as x approaches three critical values: -Infinity, 0, and +infinity
As x->Infinity, (x+1)/x^2 -> 0 and sin((x+1)/x^2) -> 0, so there is a horizontal asymptote at y=0. Same goes for x->-Infinity.
As x->0, sin((x+1)/x^2) oscilates faster and faster between -1 and 1. This is **NOT** an asymptote.
The answer is (I ONLY).
As x->Infinity, (x+1)/x^2 -> 0 and sin((x+1)/x^2) -> 0, so there is a horizontal asymptote at y=0. Same goes for x->-Infinity.
As x->0, sin((x+1)/x^2) oscilates faster and faster between -1 and 1. This is **NOT** an asymptote.
The answer is (I ONLY).
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the graph of sin x has no horizontal asymptotes leading to the first impression that neither 1 or 2 can be correct.
since the graph of sin x also has no vertical asymptotes, for a composite function you only need to consider whether or not the inner function has a vertical asymptote. In fact it does when the denominator is equal to 0 it is undefined. Thus setting x^2 = 0 and solving gives x = 0 for a vertical asymptote which is answer 3.
since the graph of sin x also has no vertical asymptotes, for a composite function you only need to consider whether or not the inner function has a vertical asymptote. In fact it does when the denominator is equal to 0 it is undefined. Thus setting x^2 = 0 and solving gives x = 0 for a vertical asymptote which is answer 3.
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you take the limit as x approaches infinity.
lim (sin((x+1)/(x^2)))
since sin is a function you can move the limit into the inside of it
sin (lim((x+1)/(x^2)))
the limit of ((x+1)/(x^2)) is zero, because it is in the form of 1/x^2
sin (0) = 0
therefore the limit as x approaches infinity of f(x) is zero,
this means that it has an asymptote at x = 0.
so III is correct.
lim (sin((x+1)/(x^2)))
since sin is a function you can move the limit into the inside of it
sin (lim((x+1)/(x^2)))
the limit of ((x+1)/(x^2)) is zero, because it is in the form of 1/x^2
sin (0) = 0
therefore the limit as x approaches infinity of f(x) is zero,
this means that it has an asymptote at x = 0.
so III is correct.
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as x → ∞, (x+1)/x² → 0, so you have a horizontal asymptote at y = 0.
as x → 0, (x+1)/x² → ∞, so sin of that whips back and forth between ±1, faster and faster as x → 0. I don't think that counts as a vertical asymptote. It IS a discontinuity, so f(x) is not differentiable at x = 0.
as x → 0, (x+1)/x² → ∞, so sin of that whips back and forth between ±1, faster and faster as x → 0. I don't think that counts as a vertical asymptote. It IS a discontinuity, so f(x) is not differentiable at x = 0.
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We can straight away eliminate the second option, as there is no reason why (x+1)/x^2 cannot equal 90 degrees (or pi/2 radians). For the first option, if y=0, (x+1/(x^2) has to equal 0. for this to happen, the top of the fraction must equal 0, thus the graph crosses the y-axis x= -1. However, x cannot equal 0, since you would end up with a divide by 0 error. Thus only the third option is true.
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I. The graph of f has a horizontal asymptote at y = 0.
limit(sin((x+1)/x^2), x = infinity) = 0
limit(sin((x+1)/x^2), x = - infinity) = 0
limit(sin((x+1)/x^2), x = infinity) = 0
limit(sin((x+1)/x^2), x = - infinity) = 0
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III.
f(θ) = sin(θ) has no asymptotes
f(x)=(x + 1) / (x^2) has an asymptote at x=0
f(θ) = sin(θ) has no asymptotes
f(x)=(x + 1) / (x^2) has an asymptote at x=0