Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 8 July 2014

Lecture 7

Limits

limnan= means an gets closer and closer to as n gets larger and larger.

limxaf(x)= means f(x) gets closer and closer to as x gets closer and closer to a.

Distance and absolute value

The absolute value on is 0 x|x| given by |x|= { x, ifx>0, 0, ifx=0, -x, ifx<0. The complex numbers is ={x+iy|x,y} withi2=-1. The absolute value on is 0 x|x| given by |x+iy|= |x2+y2|. The vector space 3 is 3= {(x,y,z)|x,y,z}. The absolute value on 3 is 30 v=(x,y,z)|v| given by |v|=|x2+y2+z2|, ifv=(x,y,z). So limn an=means limn |an-|=0, limxa f(x)=means limxa |f(x)-|=0.

Official definitions in Maths

The sequence (an) converges to if (an) satisfies if ε>0 then there exists N>0 such that
if n>0 and n>N then |an-|<ε.
Write limnan= if (an) converges to .

The function f(x) converges to as xa if f(x) satisfies: if ε>0 then there exists δ>0 such that
if |x-a|<δ then |f(x)-|<ε.

Write limxaf(x)= if f(x) converges to as xa.

A function f(x) is continuous at x=a if limxaf(x)=f(a).

MOST IMPORTANT PROPERTY of absolute value: |x+y||x| +|y|.

Useful properties of limits

(a) Assume that limnan and limnbn exist. Then
(a1) limn (an+bn) = limnan+ limnbn,
(a2) limn anbn = (limnan) (limnbn),
(a3) limn -an = -(limnan),
(a4) If (an) satisfies: if n>0 then an0, then limn1an =1limnan.
(b) Assume that limxaf(x) and limxag(x) exist. Then
(b1) limxa (f(x)+g(x)) = limxaf(x) limxag(x),
(b2) limxa (f(x)g(x)) = (limxaf(x)) (limxag(x)),
(b3) limxa (-f(x)) = -limxaf(x),
(b4) If f(x) satisfies: f(x)0 for all x close to a then limxa(1f(x))= 1limxaf(x).

(a) Assume that (an) and (bn) are sequences in and limnan and limnbn exist. If anbn then limnanlimnbn.
(a') Assume that f(x) and g(x) are real valued functions and limxaf(x) and limxag(x) exist. If f(x)g(x) then limxaf(x)limxag(x).
(b) Assume that limxag(x)= and limyf(y) exists.

Let x. Then limyf(y)=limxaf(g(x)).

(a) limnxn= { 0, if|x|<1, diverges, if|x|>1, 1, ifx=1, diverges, if|x|=1 andx1.
(b) Let n>0 and a. limxaxn=an (i.e. f(x)=xn is continuous).
(c) Let a. limxaex=ea (i.e. f(x)=ex is continuous).
(d) limn1+x+x2++xn= { 11-x, if|x|<1, diverges, if|x|1.
(e) limn(1+x+x22!++xnn!) exists.

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100315Lect7.pdf and was given on 15 March 2010.

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