Question

Prove the theorem that if limx→d h(x) = P and limx→d k(x) = Q and h(x) ≥ k(x) for all x in an open interval containing d, then P ≥ Q by using the formal definition of the limit. I know that it can be proved by contradiction assuming that P 12 years ago

Answer 1

ive seen this Qs before :/

Answer 2

because it is almost in every analysis/calculus textbook

Answer 3

aha but the prove want epsilon delta i think right ?

Answer 4

Yeah I have to prove it with the formal definition of the limit epsilon deltas. not fun

Answer 5

aha i remember now .... i tried to prove to u once..

Answer 6

http://finedrafts.com/files/CUNY/math/calculus/Spivak/Michael%20Spivak%20-%20Calculus.pdf

page 78

Answer 7

you know it says by definition, so I don't know why you're going to use by contradiction

Answer 8

I can prove it any way but I thought it would be easier by contradiction. you said page 78 right? I cant see where you are looking

Answer 9

\[\lim_{x \rightarrow \Delta x}f(x)=\frac{ f(x+\Delta x)-f(x) }{ \Delta x }\]
that is where the limit chapter starts

Answer 10

I have to prove it with the epsilon delta definition of the limit so that one does not work.

Answer 11

wats ur cours name ??

Answer 12

Math analysis the calc course after calc three

Answer 13

aha ok .. i thought that too its real analysis cours..
ok i found my old txt book at this sup(intoduction to real analysis )... i'll try to review n give u an answer.

Answer 14

your great thanks!!

Answer 15

Here is what I have right now:
Prove the theorem that if limx→d h(x) = P and limx→d k(x) = Q and h(x) ≥ k(x) for all x in an open interval containing d, then P ≥ Q by using the formal definition of the limit

Suppose that P<Q
\[\left| h(x)-P \right|<e\]
and\[\left| k(x)-Q \right|<e\]
Let \[e=\frac{ (Q-P) }{ 2 }\]

then there exist a delta (actually two of them lets choose the smaller)
such that
\[\left| x-d \right|<\delta\]

now I know that there is a contradiction because as x gets closer to h(x) or k(x) it will get further away form the limits but I do not know how to show this???

Answer 16

ok i'll continue wat u got ... and see if we get contradicton ... now im not sure abt this i might be rong k ??
lim(h(x)-k(x)=P-Q
|(h(x)-k(x))-(P_Q)|<e
h(x)-k(x) converges to (P-Q) at d from hypo. & diff
now
assume
f(x)=h(x)-g(x) ,f(x)>=0 given.
|f(x)-(P-Q)|<e
|-(P-Q)|<e......but |-(P-Q)|=|(P-Q)|
|(P-Q)|<e
-e<P-Q<e
0<P-Q+e<2e right ?
now u got
Q-P=2e
apply it
0<P-Q+e<Q-P
now u try to find a cont .... but i cant see it :/

Answer 17

@ikram002p if I plug back for epsilon I get that p-q<q-p which is what I stated I need to use the fact that (x-d)<delta and show that if epsilon is equal to (q-p)/2 that i cannot find an x if P<Q. This one really sucks it is an statement that seems easy theoretically but I cannot prove it!! Thanks for all your help

Answer 18

ur wlc... even though i dint help u that much sry...
good luck,,, and be carefull couse this course is triky trust me the limit is the esiest part :/
i hated that course .

Answer 19

Thanks again!!

Answer 20

I think I have it
Prove the theorem that if limx→d h(x) = P and limx→d k(x) = Q and h(x) ≥ k(x) for all x in an open interval containing d, then P ≥ Q by using the formal definition of the limit
Proof by contrdiction:
Assume that Q>P
\[\lim_{x \rightarrow d} h(x)-k(x) =P -Q\]
therefore for any e>0, there exist delta>0 such that
\[\left| (h(x)-k(x))-(P-Q) \right|<\epsilon \]
whenever
\[0<(x-d)<\delta\]
Since Q-P> 0 by my assumption, lets take e=Q-P
\[\left| (h(x)-k(x))-(P-Q) \right|Q-P\]
Since a <= abs(a) for any number a
\[(h(x)-f(x)) - (P-Q) < L=Q-P\]
now reducing we get
\[h(x)-k(x)<0\]
or
\[h)x)<k(x)\]
This is a contridiction since it is stated that h(x) is greater then K(x) for all x in the open interval around d. Thus
\[P \ge Q\]
QED

What do you think?

Answer 21

@ikram002p

Answer 22

ok every thing sounds right untill this one
" Since a <= abs(a) for any number a"
now see i get a contradiction
|k(x)-h(x)-(P-Q)|<Q-P
||k(x)-h(x)|-|(P-Q)||<=|k(x)-h(x)-(P-Q)|<Q-P ..but u know -|(P-Q)| sighn right ?
so

||k(x)-h(x)|+(Q-P)|<=|k(x)-h(x)-(P-Q)|<Q-P and k(x)-h(x)>0
|k(x)-h(x)+(Q-P)|<Q-P=e which is a contradiction.
any step u r confused in just ask :)