limits involving natural exponential2

[synergix]

1. The problem statement, all variables and given/known data

lim(e-2x cosx)
x-> infinity

3. The attempt at a solution

I thought right away that the limit would not exist because of cos x oscillating the function between + and - but the answer in the book says zero. I need help figuring out why my thinking was incorrect.

[Dick]

-1<=cos(x)<=1. Whether there's a limit depends on what e^(-2x) does. What does it do? For a proof think about using the squeeze theorem.

[lanedance]

think about what happens to the negative exponential as x gets large, and maybe try a squeeze theorem

[synergix]

e^(-2x) will get very small.

[synergix]

well it will still be oscillating but approaching zero all the same I suppose

[Dick]

e^(-2x) will get very small.

Ok, so what's you conclusion? And why?

[Dick]

well it will still be oscillating but approaching zero all the same I suppose

That's fine, if you don't have to prove it and omit the 'I suppose'.

[synergix]

How would I apply the squeeze theorem to this problem?

[Dick]

How would I apply the squeeze theorem to this problem?

Can you find two functions f(x) and g(x) such that f(x)<=cos(x)e^(-2x)<=g(x) such that f(x) and g(x) both approach zero? Possibly using -1<=cos(x)<=1?

[synergix]

Because
-1<=cos(x)<=1

-e-2x <= cos(x)e-2x <= e-2x

is that it? I just emulated what I have seen on a few math help sites. But it makes sense now that I have thought it out a bit.

soo

lim -e-2x=0=lim e -2x
x->infinity x->infinity

so

lim cos(x)e-2x=0
x->infinity

[Dick]

Because
-1<=cos(x)<=1

-e-2x <= cos(x)e-2x <= e-2x

is that it? I just emulated what I have seen on a few math help sites. But it makes sense now that I have thought it out a bit.

soo

lim -e-2x=0=lim e -2x
x->infinity x->infinity

so

lim cos(x)e-2x=0
x->infinity

Brilliant. A little emulation goes a long way.