Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In case of $\delta$-$\epsilon$ definition of limit, books usually show that some $L$ is the limit of some function and then they prove it by reducing $L$ from the function $f(x)- L$ and showing $n>\epsilon$ and there is some $m\geq n$ ...and so on.

But just by doing this, it doesn't seem to prove anything because I can replace $L$ by any value and still try to calculate $n$.

I also tried to calculate the limit of those functions by assuming $L$ to be a variable and equating the (example, $(n(1-L)+L)/\ldots$) $n = 0$ ...it gave me right value as limit without using L'Hopital rule.

My question is: How can I do it for equations using trigonometric identities?

To make my question more clear I am taking $f(n) = (2n+3)/(n-1)$ as an example.

Now, by the definition of limit it might have a limit, say $L$, and we need to find it. So, I use the definition of limit to find it in this way: $$| (2n+3)/(n-1) - L)| < \epsilon \implies |(n(2-L)+(L+3))/(n-1)|$$ Equating product of $n$ I will get $L=2$ , which is the limit. In a way it says that the variable of the function has to go.

The question is: how shall it be done for quadratic and higher degree equations where numerator is of higher degree than denominator and trigonometric equations?

Second thing is that if $$| (2n+3)/(n-1) - 1)| < \epsilon$$ then its $n<((\epsilon+4)/(\epsilon-1))$...so still there is $n$ I get...then why do they say in books: "prove that $L$ is limit using definition so and so..". It's incomplete to me.

Edit:Elaborated examples,If $f(x)$= $|\frac{2n+3}{n-1}$-$L$|<$\epsilon$ $=>$ $|\frac{n(2-L)+(L+3)}{(n-1)}|$ Hence $n$=$|\frac{(L+3)}{(L-2)}|$, Here L is the limit we expect to check and we can replace $n$ by 1,2,3...$N$ and it will give right result.
In case of $n$=1, as the limit shall not exist, we get absurd result as
$(3+$L$)=($L$-2)$

In case of $n$=+$\infty$, $($L$-2)=\frac{L+3}{\infty}$ which approaches to $0$, hence $L$=2. This works fine in case of higher degree equations also. But fails in case of equations that are indeterminate in nature. I will post more if I find.

Note:Please try this once on your notebook.

share|improve this question
    
Can you give an example of "can replace $L$ by any value and still calculate $n$"? –  Hagen von Eitzen Sep 11 '12 at 8:08
    
In some cases, half of the battle is even knowing the limit exists! The phrase "by definition of a limit it might have a limit" seems to indicate some uncertainty with the definitions (or, I would understand if it is just uncertainty with English.) –  rschwieb Sep 11 '12 at 13:15
    
It's good to make your edits below your OP, so that you don't completely change your old question. By the way, I think a lot of us are struggling to understand what you are talking about with trig identities. Examples would be good. –  rschwieb Sep 13 '12 at 12:07
1  
If I understand you properly, your still there is n I get is wrong: if L=1 and epsilon=1, there is no n such that |(2n+3)/(n-1)-L|<epsilon. A fortiori the requirement that (forall epsilon, exists N, forall n>N,|(2n+3)/(n-1)-L|<epsilon) fails for L=1 (and for every L different of 2). –  Did Sep 16 '12 at 7:43
1  
Likewise, for L=3, n>(6+epsilon)/(1+epsilon) does not imply that |(2n+3)/(n-1)-3|<epsilon (check your computations, I suspect you mishandled absolute values). –  Did Sep 16 '12 at 7:49

2 Answers 2

When given that a function $f$ and a number $L$, it could be possible that $L$ is a limit, or that $L$ is not a limit. If your book does not have that many typos, then we should hope for the first type :)

If you have been asked to verify that $L$ really is the limit of $f(n)$, then by definition of limits you need to check taht $|f(n)-L|$ shrinks to zero in the $\epsilon-\delta$ sense.

If, on the other hand, you are asked to find the limit, then it makes more sense to find a candidate limit, and then conduct the process I mentioned above.

For your example $f(n)=\frac{2n+3}{n-1}$, you think: "when $n$ gets very large, this is approximately $2n/n$, so I speculate that the limit is $2$." Then you can do your analysis with $|\frac{2n+3}{n-1}-2|$ and prove that it is the case.

I realize that in other cases the limit might not be so transparent. That's why we have a lot of tools (like the Squeeze Theorem) to make the limit easier to identify.

Also, once your teacher has been satisfied that you can verify a limit, you usually are allowed to "graduate" from that, and then use basic limit properties (such as "the sum of limits is the limit of the sums", and "$1/x^n$ goes to zero as $x$ goes to infinity") to perform such tasks.

To solve your example this way, a student might compute:

$\lim_{n\rightarrow\infty}\frac{2n+3}{n-1}=\lim_{n\rightarrow\infty}\frac{2n-2}{n-1}+\frac{1}{n-1}=\lim_{n\rightarrow\infty}2\frac{n-1}{n-1}+\lim_{n\rightarrow\infty}\frac{1}{n-1}=\lim_{n\rightarrow\infty}2+0=2$

That is just one of dozens of answers which are reasonable. It just depends on what your teacher is currently expecting you to do. If your teacher wants you to verify with epsilons and deltas, then do that. Ask the teacher if it's OK to find limits in this latter way, and then you might be able to do that too!

share|improve this answer
    
my whole idea is to get rid of these tools and find that one way to do things and include these tricks as special cases in it rather than mastering all tricks....I have respect for all these tools but it hides the tale. And I don't go to college, so I am free to try anything ! –  Rorschach Sep 12 '12 at 7:31
    
@rafiki That's unfortunate, because these tools "tell the tale" and offer you understanding about what's going on. If you seek a "master tool", you will find they often don't exist, and are convoluted and impractical when they do exist. And using such a tool would shelter you from the details and "hide the tale". Don't discard the laws of limits just because you think they are "tricks" :) –  rschwieb Sep 12 '12 at 12:48
    
I will take my time to respond....I want to see the reason it doesn't exist..what is the problem!! I just know one thing..these all methods are connected and special cases of each other...I wanna know how. That;s my trial. –  Rorschach Sep 12 '12 at 13:46
    
@rafiki Oh no, I don't have a proof that such a method doesn't exist. I meant to say that looking back on my own experience in getting a doctoral degree in math and teaching mathematics, I have come to that conclusion. Can you elaborate for a bit what "methods" you are talking about? I only described one method for verifying what a limit is, and one method for deducing what the limit is (without rigorous verification that it is the limit.) –  rschwieb Sep 12 '12 at 14:39
1  
@rafiki L'Hopital's rule is specifically designed to handle those cases. It's kind of like asking "why is the smallpox vaccine used against smallpox and not other diseases? It's just a special case of the general vaccine. Why won't you give me the vaccine against all diseases?" –  rschwieb Sep 13 '12 at 12:02

The idea behind the $\delta- \epsilon$ definition is to make rigorous the notion of "as $x$ gets closer to $x_0$, $f(x)$ gets closer to $L$."

For instance, suppose our function is $f(x) = \frac{x^2 - 4}{x - 2}$, and we want to talk about the limit as $x \to 2$. We can't plug this value in, but via graphing, or a table of values, etc., we might convince ourselves that as $x$ gets closer to $2$, $f(x)$ gets closer and closer to $4$. Equivalently, we could say that $|f(x) - 4|$ gets closer and closer to $0$.

This is where $\epsilon$ comes in - by "closer and closer to $0$", we mean that, for any really small positive number $\epsilon$, eventually, $|f(x) - 4| < \epsilon$. To make rigorous the phrase "eventually", we introduce $\delta$; this describes the conditions on how close $x$ needs to be to $2$ before we can be sure $|f(x) - 4| < \epsilon$. We want to find a $\delta > 0$ so that whenever $|x - 2| < \delta$ (or equivalently, $2 - \delta < x < 2 + \delta$), we have $|f(x) - 4| < \epsilon$.

In our example, if I said I want $\epsilon = 1/1000$, if you gave me $\delta = 1/10000$, that would work: whenever $1.9999 < x < 2.0001$, we can be sure $|f(x) - 4| < 1/1000$. This choice of $\delta$ isn't unique, either - you could have chosen $\delta = 1/2756$, or $\delta = 1/1001$ - any $\delta < 1/1000$ will work in this case. In this kinds of proofs, you just have to show that, given an arbitrary $\epsilon$, there's some way to find an appropriate $\delta$.

In response to your first two questions, the pattern you're observing is the rigorous definition of continuity - if a function is defined at $x_0$, and $lim_{x \to x_0}f(x) = f(x_0)$ (in this $\epsilon$-$\delta$ sense), then we say it is continuous at $x_0$. If it is continuous at every point in an interval $(a,b)$, we say it is continuous on $(a,b)$. This is the pattern you're noticing - all the functions you've been give to test are probably continuous! To see an example of a function where this doesn't hold, just try any discontinuous function - for instance $$f(x) = \begin{cases} 1 : & x \geq 0 \\ 0 : & x < 0 \end{cases}$$

share|improve this answer
    
please explain more...I don't think it answers my question or I am not able to understand ur point ! –  Rorschach Sep 11 '12 at 8:58
1  
I'm not sure what your question 3. is asking - can you elaborate or give an example of a problem you've done where you've observed this? As far as trigonometric functions, I'm not quite sure what you're asking there either, do you have an example? Are you being asked to prove a limit exists for some trig. function? –  BaronVT Sep 11 '12 at 9:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.