# help finding limit

#### zJakez

Joined Oct 2, 2006
2
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?

#### hgmjr

Joined Jan 28, 2005
9,029
Rich (BB code):
                             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

#### menouf2005

Joined Oct 4, 2006
1
what about take the transformation of cos(x)=(e(x)+e(-x))/2
so

e(9x)+e(-9x)
y=------------------
2*e(3x)

#### n9352527

Joined Oct 14, 2005
1,198
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

#### Dave

Joined Nov 17, 2003
6,970
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