Epsilon delta to N & M Definition

prasannaworld
Messages
21
Reaction score
0
Okay for a simple finite limit: e.g.
lim (3x) = 3
x->1

in the end I say:

"Therefore for every |x - 3| < delta, there exists an epsilon such that |3x-3| < epsilon"

Hence I can make delta really really small and the y bounds of epsilon will constrain the limit.



So let's come to the example I saw in an article
lim (SQRT(x)) = INF
x-INF

Okay so:
x > N - x is greater than any positive integer
Match N with M^2
x > M^2
SQRT(x) > M

Okay so how will I make my statement?
 
Mathematics news on Phys.org
so you are trying to prove that:

\lim_{x\rightarrow \infty}\sqrt{x}=\infty right?

What we want to prove is that \forall N&gt;0, \exists M&gt;0 such that whenever

x&gt;M, \sqrt{x}&gt;N


By observation we have, as you pointed out:\sqrt{x}&gt;N=&gt; x&gt;N^2

so our statement would be

\forall N&gt;0, \exists M=N^2&gt;0 such that whenever

x&gt;M=N^2=&gt;\sqrt{x}&gt;N
 
Yes. That is what I wanted. Still I think the best way for me to get this is convince myself by trying to prove a false limit (I obviously should not be able to...)
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top