Formal Definition of Limits, as x -> infinity

[kioria]

Hi,

I am having difficulties trying to adopt the formal definition of Limits as x -> infinity. I will simply try to explain my problem using an example.

The Formal Definition of Limits as x -> infinity is as follows:
Limit of f(x) as x -> infinity = L, iff we can find M such that
x > M forces absolute value of f(x) - L < e.

note: e = epsilon, M = a function of epsilon

Question:
Find as x -> infinity, the limit of: and prove using the Formal Definition.
f(x) = (sin 3x) / (x^2 + 4), the answer is 0.

Formal definition requires me to find a general equation M, in terms of e. How can I do that with these type of questions?

Here are some I can't do:

f(x) = (x + 3) / (x^2 - 3)

 

[matt grime]

So given e>0 you must show $$\frac{\sin(x)|}{x^2+4} < e$$ for all x sufficiently large. |sin(x)| is bounded above by 1, so it suffices to show that given e>0 $$\frac{1}{x^2+4}<e$$, try manipulating that expression to get some idea for M. Eg 1/x^6 <e for all x> e^{-1/6}

 

[kioria]

I have actually done that, but I am just not sure if the way I've done is right. It goes like:

$${sin(3x)} <= 1$$

$$\frac{\sin(3x)}{x^2+4} < \frac{1}{x^2+4} < \epsilon$$

$$M(\epsilon) = \sqrt{\frac{1}{\epsilon} - 4}$$

$$x = \frac{1}{\sqrt{\epsilon}} > M(\epsilon)$$

this works out so that if, I replace the x into the first condition, then it satisfies. And there I have left it as the answer.

Is that the correct approach?

 

[matt grime]

Well, x cannot be fixed by epsilon like that, M is fixed by the epsilon, and then as long as all the steps you've taken are reasonable, then x>M will imply |f(x)-L|<e

 

[kioria]

Well, x cannot be fixed by epsilon like that, M is fixed by the epsilon, and then as long as all the steps you've taken are reasonable, then x>M will imply |f(x)-L|<e

The x I meant was the x I've picked to test... Is there a better way to generalise the x I pick to test?

 

[matt grime]

you don't pick an x to test. you pick an M dependent on epsilon, and then it must be true for all x>M that..., you can't just pick one x and hope that the infinitely many other cases are also true.

 

[kioria]

you don't pick an x to test. you pick an M dependent on epsilon, and then it must be true for all x>M that..., you can't just pick one x and hope that the infinitely many other cases are also true.

And.. how would I do that? Can you show me how it is done with the above example? Thank you.

Thanks for your quick replies... really appreciate that.

 

[matt grime]

(It's better than being bored to death by the flat pitch in the test match in the Windies right now)

Let's do the easiest example: 1/x tends to zero as x tends to infinity.

Given e>0 we want 1/x<e, or x>1/e, so observing this we make the claim: let M=1/e

If x>M then 1/x<1/M and by choice 1/M=e, and we have solved the problem.

So we are given e, we manipulate the expression to see what M ought to be like, and then we check that if we assume x>M, that things behave themselves.

 

[kioria]

(It's better than being bored to death by the flat pitch in the test match in the Windies right now)

Let's do the easiest example: 1/x tends to zero as x tends to infinity.

Given e>0 we want 1/x<e, or x>1/e, so observing this we make the claim: let M=1/e

If x>M then 1/x<1/M and by choice 1/M=e, and we have solved the problem.

So we are given e, we manipulate the expression to see what M ought to be like, and then we check that if we assume x>M, that things behave themselves.

A rather, refreshing reply. Thanks for your easy example, and because of that I have much clearer understanding of this formal definition of limits. Now going back to the original example I've posted... I still don't have a clue, any more hints? Sorry to bother you... but I am just not getting at this...

 

[matt grime]

Well, we've done the manipulation so we know that letting M = sqrt(1/e-4) is sensible (and assuming e is small enough so that isnt' imaginary) then

x>M => x^2>M^2 => x^2+4> M^2+4 => (x^2+4)^{-1}< (M^2+4)^{-1} = e

so we've shown x>M implies f(x) <e

 

[kioria]

Well, we've done the manipulation so we know that letting M = sqrt(1/e-4) is sensible (and assuming e is small enough so that isnt' imaginary) then

x>M => x^2>M^2 => x^2+4> M^2+4 => (x^2+4)^{-1}< (M^2+4)^{-1} = e

so we've shown x>M implies f(x) <e

Oh my god... thanks very much for taking your time to reply. I have finally understood, and now I can do the questions that follows... Thank you.

Did you say you were watching a cricket match? GO AUSTRALIA!

 

[matt grime]

We (England) have got a chance to be the first team in history to whitewash the Windies in their own backyard. We already matched the feat of winning the first three like Oz did last year. I know, the Poms doing well, I can't believe it either. Don't worry, WI have just reached 300 for 2, sorry, 2 for 300 off 75 overs, so we won't be the ones to make history. Good on ya, (sorry, couldn't resist) for figuring it out.

 

[kioria]

Good on ya, (sorry, couldn't resist) for figuring it out.

Yes, I couldn't resist too

 

[Stevo]

(It's better than being bored to death by the flat pitch in the test match in the Windies right now)

Eh, yer just bitter you won't whitewash the series 4-0.