If we evaluate the function at infinity, we can immediately see that:

Therefore, we must perform an algebraic manipulation in order to get rid of the indeterminacy.
We can solve this limit in two ways.
<h3>Way 1:</h3>
By comparison of infinities:
We first expand the binomial squared, so we get

Note that in the numerator we get x⁴ while in the denominator we get x³ as the highest degree terms. Therefore, the degree of the numerator is greater and the limit will be \infty. Recall that when the degree of the numerator is greater, then the limit is \infty if the terms of greater degree have the same sign.
<h3>Way 2</h3>
Dividing numerator and denominator by the term of highest degree:



Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.
Answer:
y = (1/2) cos (x)
Step-by-step explanation:
It has the same period as cos x but only goes half as high and half as low. Normalyy, cos(x) goes up to 1 and down to -1.
Answers:
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Explanation:
Recall that tangent is the ratio of opposite over adjacent
tan(angle) = opposite/adjacent
So for reference angle G, we say,
tan(G) = JH/GJ = 2/1 = 2
We'll treat tan(H) in a similar fashion, but the opposite and adjacent sides swap roles. That means we'll apply the reciprocal to the result above to get 1/2 for tan(H)
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So we have this interesting property where
tan(G)*tan(H) = 2*(1/2) = 1
In general,
tan(A)*tan(B) = 1 if and only if A+B = 90 degrees
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Side note: The side sqrt(5) isn't used at all.
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Step-by-step explanation: