Expert: Paul Klarreich - 11/15/2010

Question
QUESTION: Prove: If f(x)->L as x->a, then:
If k<L, there exists delta>0 such that f(x)>k whenever 0<|x-a|<delta

ANSWER: Questioner:   Brittany
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  limits
Question:  Prove: If f(x)->L as x->a, then:
If k<L, there exists delta>0 such that f(x)>k whenever 0<|x-a|<delta
..........................................
Start with definitions:

f(x)->L as x->a  MEANS

lim[x->a] f(x) = L,  which MEANS

Given  e > 0, we can find  d > 0 such that

[e is usually called epsilon, d is delta]

| f(x) - L | < e   whenever  0 < | x - a | < d,

SO, if you have  k < L, take L - k as your e.

Then whenever 0 < | x - a | < d, you will have:

| f(x) - L | < e, which MEANS:


-e < f(x) - L  < e,  which means:

L - e < f(x) < L + e, which means:

f(x) > L - e, and since  e = L - k,

f(x) > L - (L - k)

f(x) > L - L + k

f(x) > k

That should do it.  These proofs are fun, yes??

---------- FOLLOW-UP ----------

QUESTION: prove that If f(x)->L as x->a, then:
f is bounded on some deleted neighborhood of a.  

Answer
QUESTION: prove that If f(x)->L as x->a, then:
f is bounded on some deleted neighborhood of a.
 
All you need to do is take

|x - a| < d

as your deleted neighborhood [x = a is deleted].  Then:

| f(x) - L | < e  and

- e < f(x) - L < e, so

L - e < f(x) < L + e

and that is it.