Search results
Results From The WOW.Com Content Network
This answers why epsilon, the smallest positive (surreal) number, can not be rational. The rational numbers and the real numbers have the archimedian property that between any two rationals and between any two reals there exist a third. So it is impossible to have a smallest positive real or rational number. So if a smallest positive number is ...
$\begingroup$ @user13985: The $\delta$ can't both be too large and successful. Once the $\epsilon$ gets small enough, I will be forced to pick $\delta$.
The limit definition does not define a unique value for $\delta$ in function of $\epsilon$. If you find some $\delta(\epsilon)$ that satisfies the limit definition, then any $\delta'(\epsilon)$ for which $\delta'(\epsilon)<\delta(\epsilon)$, also satisfies the limit definition.
Epsilon naught (ε₀) is the vacuum permittivity constant, representing the electric permittivity of free space. It has a value of approximately 8.85 x 10^ (-12) farads per meter. Epsilon Naut in ...
So the "epsilon" in the linked question (the entire $\min(\ldots)$ expression) can take on any value $< 1$ by choosing the appropriate $\varepsilon$. But if the "epsilon" Is given in the form $1+\varepsilon$ for $\varepsilon > 0$ then the proof is invalid.
The epsilon is given. Given an epsilon, you have to choose/find a corresponding delta. The point is that given any ϵ> 0 you have make |f(x) − f(x0)| smaller than this randomly given ϵ, by choosing/ (demonstrating the existence of) a δ> 0 so that when |x − x0| is smaller than δ, then |f(x) − f(x0)| is smaller than ϵ.
$\begingroup$ $\epsilon$ is not a magic value. It is a small number. We use the symbol to mean that that what we say about it will be true for any number no matter how small and we want some observations that are true for all small numbers but the number WE PICK is just a small number. It doesn't have any special properties of being smaller ...
Proof of sequence limit, using epsilon-delta method. (an) (a n) is a sequence with an = 1 n−−√ a n = 1 n. Proove that limn→∞an = 0 lim n → ∞ a n = 0, using epsilon-delta method. First of all, I assume that n ∈ [1; +∞) n ∈ [1; + ∞). For all n n, which are larger or equal to 1 1, nth n t h term of sequence is strictly larger ...
My first equivalence is the definition of absolute value (or if you like the case-by-case definition used by both Brian and DonAntonio, then I used both the definition and the equivalence $-(x-a)<\epsilon\iff(x-a)>-\epsilon$). $\endgroup$
Well, no, but if $\epsilon>1$ then any value $|z|<\frac12$ would also satisfy $|z|<\epsilon.$ So, effectively, for limits, you can assume $\epsilon <1$ or $0<\epsilon<\alpha$ for any real $\alpha.$ This is true because of how $\epsilon$ is used in limits. If you prove it for a particular $\epsilon_0>0,$ then it is true for all $\epsilon\geq ...