Limits involving natural exponential2

[synergix]

Homework Statement

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

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.