# Trig QWatch

This discussion is closed.
#1
Hi, I really need help with these questions.
1. Find the Fundamental period of the following periodic function

-7 cos(6 pi x )

2. A rational function is a quotient of two polynomials. Evaluate the following limit of a rational function

..lim.. -5x^9 -x^7 -8x^5 -x^4 -8x^3 -5x^2
x-->∞ 7x^7 +7x^6 +2x^5 + 2x^4 + 9x + 8
Thanks
0
14 years ago
#2
For the second one, divide the nominator by the denominator. You'll probably get something like A + B/C, and B/C will probably tend to 0 as x tends to infinity, so the whole function will tend to what A tends to.
0
14 years ago
#3
For the second, the numerator has order 9, the denominator has order 7, so the limit will tend to -infinity.
0
14 years ago
#4
(Original post by JamesF)
For the second, the numerator has order 9, the denominator has order 7, so the limit will tend to -infinity.
Yeah, that's what I wanted to say, but decided to give some explanation.
0
14 years ago
#5
(Original post by shift3)
Yeah, that's what I wanted to say, but decided to give some explanation.
Oh, sorry.

For the first, cos(x) has period 2pi, cos(ax) has period 2pi/a, so your function will have period, 2pi/6pi = 1/3.

Alternatively, for what value of a does cos(6pi(x+a)) = cos(6*pi*x)? Expanding the left,
cos(6*pi*x)cos(6*pi*a) - sin(6*pi*x)sin(6*pi*a) = cos(6*pi*x), so sin(6*pi*a) = 0 and cos(6*pi*a) = 1, hence 6*pi*a = 2pi => a = 1/3
0
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