Proving Limit Property: Easy Steps

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Homework Help Overview

The discussion revolves around proving the limit property related to derivatives, specifically the equivalence of two limit expressions involving the function f(x) as x approaches a and the variable h approaching 0.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the denominators x - a and h, with some suggesting substitutions to facilitate the proof. Others mention using a delta-epsilon argument to establish the limits.

Discussion Status

There is an ongoing exploration of the relationship between the variables involved, with some participants suggesting specific substitutions and others expressing uncertainty about the clarity of the hints provided. Guidance has been offered regarding the use of transformations in the limits.

Contextual Notes

Some participants express confusion regarding the subtleties of the problem, indicating that the discussion may involve assumptions or definitions that are not fully articulated.

andilus
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how to prove
lim(x->a)\frac{f(x)-f(a)}{x-a}=lim(h->0)\frac{f(a+h)-f(a)}{h}

it seems to be obvious, but i don't know how to prove```
 
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What is the relationship between the denominator x - a and the denominator h?
 
statdad said:
What is the relationship between the denominator x - a and the denominator h?

Sorry,I have not express clearly.
what i want to prove is just:
lim(x->a)(\frac{f(x)-f(a)}{x-a})=lim(h->0)(\frac{f(a+h)-f(a)}{h})
 
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 \epsilon > 0, there exists a \delta >0 such that for all x with the property 0 < | x - a | < \delta, then |f(x) - L | < \epsilon.

In this argument, we have the distance between a point x and a fixed point a bounded between 0 and some fixed \delta. Can you provide a similar argument as h approaches ____ ?
 
Why not let h = x - a and rewrite the limit after the substitutions?
 
That is exactly what statdad was suggesting!
 
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:
 
You still don't get it? In the first limit,
\lim_{x\to a}\frac{f(x)- f(a)}{x-a}

let h= x- a. Then x= ??
 
  • #10
wisvuze said:
use a delta-epsilon argument, the standard delta epsilon definition:

as x approaches a, we have:

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

In this argument, we have the distance between a point x and a fixed point a bounded between 0 and some fixed \delta. Can you provide a similar argument as h approaches ____ ?
I'm pretty sure none of this is applicable to the problem in this thread.
 
  • #11
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.
 
  • #12
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:

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