How to prove the limit does not exist at 0

Printable View

March 27th 2011, 02:11 PM
alice8675309

How to prove the limit does not exist at 0

I know that all of the following functions with domain R-{0}, the limit does not exist at 0. However, I'm not sure how to prove them using the Divergence Criterion: f does not have a limit at a if and only if there is an input sequence (xn) with elements in D-{a} such that (xn) converges to a by (f(xn)) diverges.

(1) f(x) = sin(1/x)
(2) f(x) =x+sin(1/x)
(3) f(x)=(1/x)sin(1/x)
March 27th 2011, 03:31 PM
RaisinBread

If you look at the first function, and pose $x_n = \frac{1}{(\frac{n \pi}{2})}$ , well you can easily prove that $x_n$ converges to 0, since it is only the harmonic sequence multiplied by a constant, and you notice that $f(x_n)=sin((\frac{n \pi}{2}))$ ={1, 0, -1, 0, 1 ...}

It can easily be shown that f(xn) diverges, since there exists an epsilon greater than zero(just pick smaller than 1) for which, no matter what rank N of the sequence $f(x_n)$ you pick, there exists i greater than N such that $f(x_i)$ - $f(x_{i+1})$ is greater than epsilon(two successive terms in f(xn) always have a distance of exactly 1), therefore $f(x_n)$ is not Cauchy, which implies that it does not converge.
March 27th 2011, 06:46 PM
alice8675309

Quote:

Originally Posted by RaisinBread

If you look at the first function, and pose $x_n = \frac{1}{(\frac{n \pi}{2})}$ , well you can easily prove that $x_n$ converges to 0, since it is only the harmonic sequence multiplied by a constant, and you notice that $f(x_n)=sin((\frac{n \pi}{2}))$ ={1, 0, -1, 0, 1 ...}

It can easily be shown that f(xn) diverges, since there exists an epsilon greater than zero(just pick smaller than 1) for which, no matter what rank N of the sequence $f(x_n)$ you pick, there exists i greater than N such that $f(x_i)$ - $f(x_{i+1})$ is greater than epsilon(two successive terms in f(xn) always have a distance of exactly 1), therefore $f(x_n)$ is not Cauchy, which implies that it does not converge.

ok so say for the second one, would you use n+1/n and show that that diverges. Sorry im just trying to work in that divergence criterion. For the last one, would I show something along the lines of (1/x)(1/x)?

Also, don't these proofs need epsilon and delta? To use the Divergence Criterion?
March 28th 2011, 05:36 AM
HallsofIvy

YOU stated, in your first post that the Divergence Criterion say s " f(x) does not have a limit at a if and only if there is an input sequence (xn) with elements in D-{a} such that (xn) converges to a by (f(xn)) diverges." It is not always necessary to use "epsilon-delta" proofs to show that a sequence does or does not converge.
March 28th 2011, 05:44 AM
alice8675309

Quote:

Originally Posted by HallsofIvy

YOU stated, in your first post that the Divergence Criterion say s " f(x) does not have a limit at a if and only if there is an input sequence (xn) with elements in D-{a} such that (xn) converges to a by (f(xn)) diverges." It is not always necessary to use "epsilon-delta" proofs to show that a sequence does or does not converge.

Oh, ok so basically I just have to show the imput sequence and show it converges but that f(xn) diverges?
March 28th 2011, 12:40 PM
RaisinBread

Yes, and the input sequence has to converge to zero in your case. Maybe your confusion comes from the fact that, the definition of a convergent function involves epsilon-delta.

However this criterion does not. A more intuitive way of explain it is that, if there is "some way" of approaching zero (in our case the input sequence xn) such that f(xn) does not converge, this must mean that f(x) doesn't converge at x=0, because if it were the case, it would converge no matter how you approach zero.

as for using n + 1/n, if you mean using that as an input sequence for the second function, I don't think it would work, because your input sequence has to converge to zero, otherwise you can't use the criterion you mentioned in the first post. For the criterion to work, your input sequence MUST converge to zero, and the f(xn) diverge.