2
$\begingroup$

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.

  • 0
    Can you give an example of "can replace $L$ by any value and still calculate $n$"?2012-09-11
  • 0
    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.)2012-09-11
  • 0
    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.2012-09-13
  • 0
    The paragraph where you arrive at the condition $n\lt(\epsilon+4)/(\epsilon+1)$ is wrong, due to some basic algebra errors. You might try to compute again what the condition $|(2n+3)/(n-1)-1|\lt\epsilon$ means.2012-09-15
  • 0
    @did:sorry it was $\epsilon$-1 and I wrote it +1...thanks for correcting.2012-09-15
  • 0
    But surely you noted that at least for small enough epsilon, this condition is never met for positive values of n, hence your next sentence should probably be modified (and this indicates exactly why (2n+3)/(n-1) does not converge to 1).2012-09-15
  • 0
    @did:Actually my question is that if n can be calculated for all values greater than $L$, than how does the definition help ! For example in case of $L$=3, $n$>$\frac{\epsilon+6}{\epsilon+1}$. By looking at definition I cannot differentiate between $2$ and $3$2012-09-16
  • 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|N,|(2n+3)/(n-1)-L|2012-09-16
  • 1
    Likewise, for L=3, n>(6+epsilon)/(1+epsilon) **does not imply** that |(2n+3)/(n-1)-3|2012-09-16
  • 0
    @did: "for all epsilon"..thanks for this reminder..I was silly to miss this one, but in books they never give such examples where this can be known.So, this to be true in a way the reminder should always be just $\epsilon$ rather than anything else..!!2012-09-16

2 Answers 2