• Support PF! Buy your school textbooks, materials and every day products Here!

Basic limit proof of limit equivalence

  • Thread starter a_skier
  • Start date
  • #1
17
0

Homework Statement



Prove that limx[itex]\rightarrow[/itex]cf(x)=L if and only if limh[itex]\rightarrow[/itex]0f(x+h)=L.

Homework Equations





The Attempt at a Solution



I think this is a simple problem, but I am getting caught up in the middle, as I'm not sure if my procedure is a valid way to prove the statement.

Suppose limx[itex]\rightarrow[/itex]cf(x)=L.

Then for each [itex]\epsilon[/itex]>0 there is a [itex]\delta[/itex]>0 so that

if |x-c|<[itex]\delta[/itex] then |f(x)-L|<[itex]\epsilon[/itex].

Let x=c+h. Then if |x-c|<δ [itex]\Rightarrow[/itex] |c+h-c|<δ [itex]\Rightarrow[/itex] |h|<δ. Furthermore, |f(c+h)-L|<ε. And since we assumed limx[itex]\rightarrow[/itex]cf(x)=L, it follows then that limh[itex]\rightarrow[/itex]0f(c+h)=L.*

So my questions is this: it is valid to simply let x=c+h as I did?

*Note, I did not prove the other case, but I just wrote this out so you guys can get a good idea of what my argument is and tell me if it is wrong or not.
 

Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
This is the right approach, but your arrows don't point the way you want them to. To show that if h goes to zero, f(c+h) goes to L, you need to start with: if |h| < δ, things happen. You started by assuming that |x-c| < δ, then showed for your choice of defining h that |h| < δ as well, which isn't good enough. But all the stuff you wrote down is reversible, so you should be able to start with |h| < δ and then use what you know about f(x) when x=c+h.
 
  • #3
17
0
Thank you very much Office_Shredder! One question though: I thought that I had to prove it in both orders, since it is an if and only if statement. Your route certainly seems more sensible but if I were to write the complete solution, wouldn't I assume in one case that as h goes to zero f(c+h) goes to f(c) and show that f(x) going to f(c) follows, and then start over and assume that as x goes to c f(x) goes to f(c) and show that f(c+h) going to f(c) follows from there? What I am asking is if I wanted to prove A iff B, wouldn't I do:

If A is true, then B follows.
If B is true, then A follows.

Thus, A is true iff B is true. ? (Hopefully that example makes more sense than the above).

Again, thank you!
 
  • #4
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
Yes, but it's important to prove if A then B in the section where you are claiming if A then B. If you want to assume
[tex] \lim_{x\to c} f(x) = L [/tex]
and then prove that
[tex] \lim_{h\to 0} f(c+h) = L [/tex],
then you better at some point have h be an arbitrary number such that |h| < δ
 
  • #5
17
0
Ohhhhh! I see. Thank you!
 

Related Threads on Basic limit proof of limit equivalence

  • Last Post
Replies
1
Views
1K
Replies
12
Views
773
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
12
Views
5K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
7K
  • Last Post
Replies
1
Views
985
Top