squeeze rule

Let f,g,h: be three sequences of real numbers such that


for all n. If limnf(n) and limnh(n) exist and are equal, say to a, then limng(n) also exists and equals a.

The proof is fairly straightforward. Let ϵ be any real number >0. By hypothesisMathworldPlanetmathPlanetmath there exist M,N such that

|a-f(n)|<ϵ for all nM
|a-h(n)|<ϵ for all nN

Write L=max(M,N). For nL we have

  • if g(n)a:

  • else g(n)<a and:


So, for all nL, we have |g(n)-a|<ϵ, which is the desired conclusionMathworldPlanetmath.

Squeeze rule for functions

Let f,g,h:S be three real-valued functions on a neighbourhood S of a real number b, such that


for all xS-{b}. If limxbf(x) and limxbh(x) exist and are equal, say to a, then limxbg(x) also exists and equals a.

Again let ϵ be an arbitrary positive real number. Find positive reals α and β such that

|a-f(x)|<ϵ whenever 0<|b-x|<α
|a-h(x)|<ϵ whenever 0<|b-x|<β

Write δ=min(α,β). Now, for any x such that |b-x|<δ, we have

  • if g(x)a:

  • else g(x)<a and:


and we are done.

Title squeeze rule
Canonical name SqueezeRule
Date of creation 2013-03-22 13:46:31
Last modified on 2013-03-22 13:46:31
Owner Daume (40)
Last modified by Daume (40)
Numerical id 4
Author Daume (40)
Entry type TheoremMathworldPlanetmath
Classification msc 26A03
Synonym squeeze theorem
Synonym squeeze test