Correct this improper definition of a limit

  • Thread starter Thread starter Easy_as_Pi
  • Start date Start date
  • Tags Tags
    Definition Limit
Easy_as_Pi
Messages
31
Reaction score
0

Homework Statement


Eddy wrote on his midterm exam that the definition of the limit is the
following: The sequence {an} converges to the real number L if there
exists an N ∈ Natural numbers so that for every \epsilon > 0 we have |an − L| < \epsilon for all
n > N. Show Eddy why he is wrong by demonstrating that if this were
the definition of the limit then it would not be true that lim n→∞ 1/n = 0.
(Hint: What does it mean if |a − b| < \epsilon for every \epsilon > 0?)


Homework Equations


|a-b| <ε means that ||a|-|b|| < ε from the reverse triangle inequality


The Attempt at a Solution


I know it has to do with the fact that the actual definition of a limit has "for every ε > 0, there exists an N \in Natural numbers S.T. ..." so, Eddy reversed that part of the definition. I just haven't been able to quite see the difference of the two. A little push in the right direction would be greatly appreciated. I like figuring these out on my own, so no full on answers, please.
 
Physics news on Phys.org
So there exists a certain N.

Let \varepsilon = |1/N|[/tex]<br /> <br /> and try to prove that 1/n does not converge to 0 with this definition.
 
I'm not quite sure I follow. I ended up answering it this way: Eddy's definition implies there is a single natural number, N, such that for all n>N |1/n|< every epsilon greater than zero. Which is not true. For every epsilon you give me, I can find an N such that 1/n is less than that epsilon for all n>N, but if you pick a newer, smaller epsilon, my N has to be larger, and since the natural numbers are unbounded, we can do this forever. But, it's a different N for each new epsilon, not one single N like eddy implied. Does that make sense?
 
That's right.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top