Question

Can someone explain to me this equation? lim x-->infinity ((e^-3x)cos(x))

Answer 1

What can you say about the cosine as x increases without bound?

Answer 2

essentially I know that cos(x) = 0
and e^-3x*0 = 0
I feel like there has to be more and I am missing it

Answer 3

No, cosine doesn't do that with increasing x. Try again.

Answer 4

you can look at the graph of e^-x

as x gets large, that value becomes smaller, 1/e^x , for ever

Answer 5

the trig functions are periodic, cosine bounces back and forth between 1 and -1 forever as x, or the angle becomes larger , to infinity

Answer 6

Too bad. Dan spilled the beans. The cosine does not help us, but it is sell enough behaved. If the exponential cooperates, which it does, we can still pin this one down.

Answer 7

oh, sorry, not too much, ,just suggest graphing the g(x)*f(x) both f and g seperate and look what they do for large values

Answer 8

-1 <+ cos(x) <=1
-e^(-3x)<=e^(-3x) = 0
e^(-3x)*cos(x) =0

Answer 9

-1<= cos(x) <=1

Answer 10

so cos(x) is going in between -1 and 1

Answer 11

You can use the definition of limit to show this exists prolly, delta-epsilon

or squeeze theorem, , since cosx) is only 1 to -1

-e^(-x) <= e^(-x)*cos(x) <= +e^(-x)

both left and right limits are 0, so the middle prolly is too

Answer 12

Thank you