help; find limit of (e^-3x)(cos9x)

[zJakez]

hi, i need help solving this limit equation.

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:

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)

Now, I am stuck at this point. Am I approaching this equation correctly?
Can the answer be 0? Since on the graphing calculator, x-->infinity is closer to 0, therefore the answer must be 0, but I'm not sure.
Are there other possible ways to solve this equation?

thks alot.

 

[skeeter]

let's use some common sense here ...

e^-3x*cos(9x) = cos(9x)/e^3x

as x->infinity, the numerator cos(9x) oscillates between the finite values -1 and 1.

as x->infinity, the denominator e^3x grows without bound.

so, the fraction ...
(something that stays between 1 and -1)/(somewthing that gets really big really fast)
should approach what value?

 

[zJakez]

hmm ok, from what i got from you:

(something that stays between 1 and -1) which is 0

/ (divide)

(somewthing that gets really big really fast) which is infinity

then the value = 0, right?

 

[tkhunny]

(something that stays between 1 and -1) which is 0

Run that by me again? Think about that real hard and tell me why it is utterly incorrect.

Despite that error, what you meant to say was that it is FINITE or it has a BOUND, while the denominator is UNbounded.

You DO have to practice the language. Now, tell my what's wrong with that statement I quoted.

 

[zJakez]

What is it then?
I still don't see any # between 1 and -1 besides 0.
Why is it wrong?

 

[pka]

The following is TRUE:
\(\displaystyle \lim _{x \to \infty } \frac{1}{x} = \lim _{x \to \infty } \frac{{10000}}{x} = \lim _{x \to \infty } \frac{{\sin (x)}}{x} = \lim _{x \to \infty } \frac{{B(x)}}{x} = 0\)

Where B(x) is any bounded function.
Do you know what a bounded finction is?

 

[skeeter]

What is it then?
I still don't see any # between 1 and -1 besides 0.
Why is it wrong?

the only integer between 1 and -1 is 0.

however, the value of cos(9x) is not restricted to integer values, it takes on all real values between -1 and 1 ... not just 0, correct?

 

[tkhunny]

What is it then?
I still don't see any # between 1 and -1 besides 0.
Why is it wrong?

If I owe you $0.50, I can just keep it because you don't think it exists?

Cool!

 

[zJakez]

The following is TRUE:
\(\displaystyle \lim _{x \to \infty } \frac{1}{x} = \lim _{x \to \infty } \frac{{10000}}{x} = \lim _{x \to \infty } \frac{{\sin (x)}}{x} = \lim _{x \to \infty } \frac{{B(x)}}{x} = 0\)

Where B(x) is any bounded function.
Do you know what a bounded finction is?

I'm not familiar with names, but I understood the equation you wrote. Something small over big is always 0.

What is it then?
I still don't see any # between 1 and -1 besides 0.
Why is it wrong?

the only integer between 1 and -1 is 0.

however, the value of cos(9x) is not restricted to integer values, it takes on all real values between -1 and 1 ... not just 0, correct?

Wait, -1 and 1, those are y numbers on the graph? The cosine graph is a wave form that goes from up 1 to down -1. Is that what you are talking about?
I still don't see why it matters to the equation that i am having a problem on.

 

[tkhunny]

It matters to keep your thinking fundamentally sound so as not to cause additional confusion in the future. Keep it as straight as you can as often and as consistently as you can. You will regret sloppiness.

 

[zJakez]

I am been kept it consistently ordered and trying to keep it as simple as it can with this equation, but u guys just kept went on about examples that are nearly unrelated to the equation. Just get straight to the point. How do I solve this equation?
Do I take derivative of it, use L'hospital Rule, or any other methods?

 

[galactus]

It appears it's undefined

\(\displaystyle \L\\\lim_{x\to\infty}e^{-3x}cos(9x)\)

\(\displaystyle \L\\e^{-3\lim_{x\to\infty}x}cos(9\lim_{x\to\infty}x)\)

\(\displaystyle \L\\\frac{cos(9\lim_{x\to\infty}x)}{\frac{1}{e^{-\infty}}\)

 

[zJakez]

It appears it's undefined

\(\displaystyle \L\\\lim_{x\to\infty}e^{-3x}cos(9x)\)

\(\displaystyle \L\\e^{-3\lim_{x\to\infty}x}cos(9\lim_{x\to\infty}x)\)

\(\displaystyle \L\\\frac{cos(9\lim_{x\to\infty}x)}{\frac{1}{e^{-\infty}}\)

thank you