Seemingly Simple Derivative (as a limit) Problem

  • Context: Graduate 
  • Thread starter Thread starter luke8ball
  • Start date Start date
  • Tags Tags
    Derivative Limit
Click For Summary

Discussion Overview

The discussion revolves around a limit problem involving derivatives, specifically the expression lim [f(ax)-f(bx)]/x as x approaches 0. Participants are exploring the application of derivative definitions and techniques to manipulate the expression.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in showing the limit and questions whether they are missing something fundamental.
  • Another participant suggests adding zero in the form of f(0)-f(0) to the numerator to facilitate rearrangement.
  • A subsequent reply proposes a rearrangement of the limit into two separate limits involving f(ax) and f(bx) evaluated at 0.
  • Another participant hints at using the chain rule as a potential approach to the problem.
  • A further hint introduces a substitution with g(x)=ax, indicating that g(0)=0, which may aid in the analysis.

Areas of Agreement / Disagreement

Participants are exploring various approaches to the limit problem, but there is no consensus on the correct method or solution as of yet.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in manipulating the limit expression, and there may be dependencies on the properties of the function f and its derivatives.

luke8ball
Messages
22
Reaction score
0
I'm having trouble showing the following:

lim [f(ax)-f(bx)]/x = f'(0)(a-b)
x→0

I feel like this should be really easy, but am I missing something? I tried to use the definition of the derivative, but I know I can't just say f(ax)-f(bx) = (a-b)f(x).

Any ideas?
 
Physics news on Phys.org
Try to add zero in your numerator in the shape f(0)-f(0), and see if you can rearrange it in a clever manner.
 
You mean so that I get:

[lim f(ax) - f(0)]/x - [lim f(bx) - f(0)]/x
x→0 x→0

I had thought about that, but I still don't see how that gives me af'(0) - bf'(0)...
 
Think chain rule..
 
A further hint:
Let g(x)=ax. Then, g(0)=0
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K