Proving Limit Property: Easy Steps

  • Thread starter Thread starter andilus
  • Start date Start date
  • Tags Tags
    Limit
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
11 replies · 2K views
andilus
Messages
8
Reaction score
0
how to prove
lim(x->a)[tex]\frac{f(x)-f(a)}{x-a}[/tex]=lim(h->0)[tex]\frac{f(a+h)-f(a)}{h}[/tex]

it seems to be obvious, but i don't know how to prove```
 
Physics news on Phys.org
statdad said:
What is the relationship between the denominator [tex]x - a[/tex] and the denominator [tex]h[/tex]?

Sorry,I have not express clearly.
what i want to prove is just:
lim(x->a)[tex](\frac{f(x)-f(a)}{x-a})[/tex]=lim(h->0)[tex](\frac{f(a+h)-f(a)}{h})[/tex]
 
Listen to what statdad is saying. What is the relationship by h and x - a? The answer to this question will essentially answer your question.
 
use a delta-epsilon argument, the standard delta epsilon definition:

as x approaches a, we have:

given [tex]\epsilon[/tex] > 0, there exists a [tex]\delta[/tex] >0 such that for all x with the property 0 < | x - a | < [tex]\delta[/tex], then |f(x) - L | < [tex]\epsilon[/tex].

In this argument, we have the distance between a point x and a fixed point a bounded between 0 and some fixed [tex]\delta[/tex]. Can you provide a similar argument as h approaches ____ ?
 
Why not let h = x - a and rewrite the limit after the substitutions?
 
I wasn't sure that that was what statdad was getting at, which is why I posted the equation. Some things are just too subtle, at least for me. :blushing:
 
wisvuze said:
use a delta-epsilon argument, the standard delta epsilon definition:

as x approaches a, we have:

given [tex]\epsilon[/tex] > 0, there exists a [tex]\delta[/tex] >0 such that for all x with the property 0 < | x - a | < [tex]\delta[/tex], then |f(x) - L | < [tex]\epsilon[/tex].

In this argument, we have the distance between a point x and a fixed point a bounded between 0 and some fixed [tex]\delta[/tex]. Can you provide a similar argument as h approaches ____ ?
I'm pretty sure none of this is applicable to the problem in this thread.
 
No, I understand perfectly how to use h = x - a to transform the first to the second; I just didn't see that statdad was hinting at doing it that way.
 
Sorry for any confusion I caused. I've always believed the best horror movies ( and books) are the ones that hint at the source of the horror, and that the best hints are ones that make you puzzle out their meaning. This time, apparently, I was a little too vague.
 
Last edited: