I don't know if I'm allow to ask calculus questions on this forum, but here goes: Find the limit of (e^-3x)(cos9x) when x--->infinity. State whether the limit is (some numbers), negative infinity/positive infinity, does not exist, or use any possible ways to find the limit. My calculations: Method1: I put it in the calculator and used the data table to replace x with any positive numbers since x--> +infinity. The answer is -infinity, since x replace with any positive numbers the result will always came out negative. Method2: Get the equation out of the limit and make it an equation: y = (e^-3x)(cos9x) Then I used the derivative of it, and since it multiplies I have to use the Product Rules. y = (-3e^-3x)(cos9x) + (e^-3x)(-9sin9x) = 3e^-3x(-cos9x - 3sin9x) It is not the answer but my teach would accept an equation written out with limit. Am I approaching this equation correctly?
Code ( (Unknown Language)): cos(9x) y = (e^(-3x))(cos(9x)) = -------- e^(3x) I think your equation can be rewritten in the form above. If I haven't goofed something up then as x approaches positive infinity it looks like the denominator will approach infinity. That would mean that as x ---> infinity, y would approach 0. The numerator is nothing more than a sinusoidal waveform with amplitude 1 and a period of 0.22pi. hgmjr
what about take the transformation of cos(x)=(e(x)+e(-x))/2 so e(9x)+e(-9x) y=------------------ 2*e(3x)
As far as I can remember, cos(x) is not equal to (e^(x) + e^(-x))/2. The identities are as follow: cos(x) = (e^(ix) + e^(-ix))/2 and cosh(x) = (e^(x) + e^(-x))/2
n9352527 is correct here. Consider: cos(jx) = cosh(x) We know: cos(x) = (e^(jx) + e^(-jx))/2 Let x = jx: cos(jx) = (e^(jjx) + e^(-jjx))/2 And j*j = -1 Therefore: cos(jx) = (e^(x) + e^(-x))/2 = cosh(x) You can do the same for sin(x) and sinh(x), from which you can get tanh(x) = sinh(x)/cosh(x) Dave