Limit Duality Theorem

Duality Theorem

In this video, I use a neat little trick to show that the limit as n goes to infinity of 2^n is infinity, by using the fact (shown before) that the limit of (1/2)^n is 0.


Other examples of limits can be seen in the playlist below.

[tomesval]

It's nice to see someone teaching about this useful math concepts in analysis.
Congrats Dr.Peyam.

[itaisilverberg]

1:46
We can show, using induction, that 2ⁿ > n for all natural n. Then, for every M>0 we choose n := ceil(M), and we have 2ⁿ > n ≥ M. Q.E.D

[LetsLearnNemo]

As always, good content. Using properties of functional inverses, reciprocals, and overall transformations for evaluating limits is such an overlooked method for dealing with limits in undergrad school. Being introduced to the idea of "u-substitution" (which is just a transformation of the space we are working with) is sometimes a very powerful and fascinating approach to solving problems (laplace transforming derivatives into polynomials is a beautiful example, imo).

[aadityajha]

Nice, you are a great teacher. Keep smiling.

[sunainapati]

Thankyou for posting! I am actually following your lecture notes to learn real analysis, really helpful :)

[dmitrypetrov]

Yes, it's trivial. Very sad that many people can't think in epsilon-language.

[jeremy.N]

how about the limit as n goes to infinity of n+1, or is that trivial?

[mohammedal-haddad]

I am curious about proving 2^n is infinite without logarithms and duality.

[iabervon]

Can't you just prove that 2^n > n for all natural numbers n [*], and then N = M is sufficient even if it's extreme overkill? It's only when you want to prove that a^n goes to infinity for all a > 1 that you want logs to find an N that will work. [*] E.g., by induction; 2^1>1, and 2^(n+1)=2^n+2^n>n+1 if 2^n>n and 2^n>1.

[hassanalihusseini]

I think recently mathematicians discovered an epsilon that is so small that if you divide it by two it becomes negative.