# Limit as x goes to infinity of e^(-x)* sin(x)

## Homework Statement

I am trying to take the following limit

lim as x goes to infinity of ( e^-x )*sin(x)

## The Attempt at a Solution

Can I say that it ges to '0' just because the 1/e^x goes to '0'. Or there is a better way to solve it?

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vela
Staff Emeritus
Homework Helper

It depends on how rigorous you need to be, but it goes to 0 essentially for the reason you said.

Is there any other reason why it goes to zero?

you can't compute a limit of a product based solely on the limit of one of the factors. use the squeeze theorem.

vela
Staff Emeritus
Homework Helper

No. It's just that if this problem is for a math class and you're covering limits, you might be expected to justify your answer better.

so if i split the 2 function apart....

the e^-x approach '0'

but the sin(x) approach nothing it will keep going from -1 to 1

right?

vela
Staff Emeritus
Homework Helper

Yes.

Mark44
Mentor

I think that the best approach is one that ice109 suggested earlier - the squeeze theorem. Note that e-x = 1/ex.

For all real x, -1 <= sin(x) <= 1
so, also for all real x, -1/ex <= sin(x)/ex <= 1/ex

The leftmost and rightmost expressions approach zero as x approaches infinity, squeezing the expression in the middle. Its limit is likewise zero.

No. It's just that if this problem is for a math class and you're covering limits, you might be expected to justify your answer better.
i have no clue what you're talking about. what you stated isn't simply unrigorous - it's completely wrong

if f(x)->0 that doesn't imply that f(x)g(x)->0.

take f(x)= 1/x and g(x)=sin(x) or f(x)=x and g(x) = exp(1/x)

for what you said to be true both limits have to exist and be finite

vela
Staff Emeritus
Homework Helper

i have no clue what you're talking about. what you stated isn't simply unrigorous - it's completely wrong

if f(x)->0 that doesn't imply that f(x)g(x)->0.
That's not what I said. For this particular problem, the exponential does cause the limit to be zero. It's just an example of the typical tendency of a decaying exponential factor to drive a function to a limit zero.

That's not what I said. For this particular problem, the exponential does cause the limit to be zero. It's just an example of the typical tendency of a decaying exponential factor to drive a function to a limit zero.
that is what you implied.

but fine - what is the limit of f(x)g(x) when x->0 if

f(x) = e^-x and g(x)= exp(1/x) ?

vela
Staff Emeritus
Homework Helper

that is what you implied.
No, it's what you inferred.

but fine - what is the limit of f(x)g(x) when x->0 if

f(x) = e^-x and g(x)= exp(1/x) ?
I knew you would respond like this and therefore questioned whether I should bother responding to you at all in the first place. Yes, I could and should have been clearer, but I'm not really looking to get into a pissing contest with you about what I meant by what I wrote.