Proving Limits at Infinity Using the Formal DefinitionDate: 04/11/2004 at 09:20:40 From: David Subject: Calculus - Formal Definition of Limits as x -> infinity. We are required to find a limit (informally), then prove that our answer is correct by using the formal definition of limits, which is "Limit of f(x) as x -> infinity is equal to L, iff we can find M (a function of epsilon) such that x > M forces |f(x) - L| < epsilon." Specific examples: a) f(x) = (sin 3x)/(x^2 + 4) b) f(x) = (x + 3)/(x^2 - 3) I do get the idea of epsilon and M, but when it comes to using them to prove, I really don't know where to start. How can I find M (a function of epsilon), and if I do find M, which x do I replace in to check? |(sin 3x)/(x^2 + 4) - 0| < epsilon so, find M by equating to get |(sin 3x)/(x^2 + 4)| = epsilon. From here I am not able to proceed. Date: 04/11/2004 at 11:24:04 From: Doctor Fenton Subject: Re: Calculus - Formal Definition of Limits as x -> infinity. Hi David, Thanks for writing to Dr. Math. With limits, "equating" the two sides usually doesn't help much, because you usually get an equation which you can't solve analytically. It is usually better to use estimation and work with inequalities. For your first limit sin(3x) lim ------- x->oo x^2 + 4 it appears that you have correctly evaluated the limit, since for very large x, the numerator is a number between -1 and +1, while the denominator is an extremely large number, so the quotient will get smaller and smaller as the denominator gets larger and larger. Formally, you have to show that given a positive epsilon (E), you can find an M such that if x > M, then | sin(3x) | | -------- | < E | x^2 + 4 | Because the left side is a very complicated expression, let's look for a simpler expression which is always larger than the complicated left side. If we increase the magnitude of the numerator of a fraction, the magnitude of the fraction increases, and if we decrease the magnitude of the denominator of the fraction, again the magnitude of the fraction increases. We know that |sin(3x)| <= 1, and |x^2 + 4| > x^2, so | sin(3x) | 1 | ------- | < --- | x^2 + 4 | x^2 So, to make the left side < E, it is sufficient to make the right side, which is much simpler to analyze, less than E. Can you now see how to choose M so that if x > M, then 1/x^2 < E? The second problem will be similar, but you need to find something larger than x + 3, so you might use x + 3 < 2x , for x > 3 (so that you will need to make sure M >= 3) and you need something smaller than x^2 - 3, so you could use x^2 x^2 x^2 - 3 > --- , as long as 3 < --- or x > sqrt(6) 2 2 If x > 3, to make the first inequality work, then x > sqrt(6), so this second inequality will also be true. If you have any questions or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ Date: 04/11/2004 at 11:40:21 From: David Subject: Calculus - Formal Definition of Limits as x -> infinity. Thanks for your reply. I have done the examples according to your suggestions, and have completed the set. Here are my answers to the second example; could you please confirm that I am on the right track with your suggestions? Thank you. |x + 3| |4x| --------- < ---------- < E (for x > 3, and M >= 3 for x > 3) |x^2 - 3| |x^2| 4 --- < E x thus: 4 --- = M(E) E and if, x > M 1 1 - < - x M 4 4 - < - x M which implies that | f(x) - L | < E, for x > M. Date: 04/11/2004 at 12:05:56 From: Doctor Fenton Subject: Re: Calculus - Formal Definition of Limits as x -> infinity. Hi David, That looks good. Usually, to indicate that you want M >= 3 (to make the estimating inequalities valid when x > M), and also M >= 4/E (to make the simplified expression less than E), you write M = max(3,4/E). Then, if x > M, the inequality |x + 3| |4x| 4 --------- < ---------- = --- |x^2 - 3| |x^2| x is true, and since x >= 4/E, then 4/x < E, so |x + 3| --------- < E |x^2 - 3| - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ Date: 04/11/2004 at 12:08:07 From: David Subject: Thank you (Calculus - Formal Definition of Limits as x -> infinity.) Thanks very much for your reply. It was very helpful, and now I feel much better! Thanks again :D |
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