Limit similar to differentiation

Click For Summary
SUMMARY

The limit of the expression (f(x + ah) - f(x + bh))/h as h approaches 0 can be evaluated using the definition of the derivative. By substituting specific functions such as x^2 and 1/2x, the results yield 2x(a - b) and 1/2(a - b) respectively. This leads to the conclusion that the limit simplifies to f'(x) * (a - b) as h approaches 0. The critical insight is recognizing that both ah and bh approach 0, allowing the application of the derivative at point x.

PREREQUISITES
  • Differentiable functions and their properties
  • Understanding of limits in calculus
  • Knowledge of the definition of the derivative
  • Basic algebraic manipulation of functions
NEXT STEPS
  • Study the formal definition of the derivative in calculus
  • Explore the concept of limits in more depth, focusing on epsilon-delta definitions
  • Learn about higher-order derivatives and their applications
  • Investigate the Mean Value Theorem and its implications for differentiable functions
USEFUL FOR

Students studying calculus, mathematics educators, and anyone interested in understanding the properties of differentiable functions and limits.

Diffy
Messages
441
Reaction score
0

Homework Statement



f a differentiable real valued function

lim h - > 0 of (f(x + ah) - f(x + bh))/h

where a,b real numbers

Homework Equations



definition of derivative

lim h-> 0 of (f(x+h) - f(x))/h

The Attempt at a Solution



I've picked several functions like x^2 and 1/2x

in the first case you get 2xa - 2xb = 2x(a-b)

in the latter you get 1/2a - 1/2b = 1/2(a-b)

This leads me to suspect f'(x)*(a - b) as the solution, I just have no idea what to do to the limit.

I tried working backwards from my solution, but I still am missing something. Any help is appreciated.
 
Physics news on Phys.org
See if this gets you anywhere.

[tex] \frac{f(x+ah)-f(x+bh)}{h} = \frac{f(x+ah)-f(x)}{h} + \frac{f(x) - f(x+bh)}{h} = a\left(\frac{f(x+ah)-f(x)}{ah}\right) - b \left(\frac{f(x) - f(x+bh)}{bh}\right)[/tex]
 
thanks, that last one since you are multiplying by -b the numerator should be f(x +bh) - f(x)

It is clear that as h -> 0 ah and bh ->0 and therefore those we are left with f'(x)

Thank you so much for your help.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 13 ·
Replies
13
Views
4K