Limit problem using the formal definition of a derivative

Click For Summary
SUMMARY

The discussion centers on determining the existence of the derivative f'(0) for the function f(x) = (x^2)sin(1/x) when x ≠ 0 and f(0) = 0. The derivative is calculated using the limit definition f'(x) = (f(x+h) - f(x)) / h. The user is advised to simplify the expression by substituting x = 0 directly into the limit, focusing on the behavior of the function as h approaches 0.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with the definition of a derivative
  • Knowledge of trigonometric functions and their properties
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the limit definition of a derivative in depth
  • Learn about the behavior of oscillating functions near points of discontinuity
  • Explore the concept of continuity and differentiability
  • Practice simplifying complex expressions involving limits
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and limits, as well as educators looking for examples of derivative calculations using formal definitions.

oates151
Messages
11
Reaction score
0

Homework Statement



Use the definition of the derivative to determine if f'(0) exists for the function,

f(x) = (x^2)sin(1/x) if x is not 0
0 if x is = 0

Homework Equations



f'(x) = f(x+h) - f(x)
------------
h

The Attempt at a Solution



Starting plugging it all in as usual and got to

(x^2 +2xh + h^2)(sin(1/x+h)) - (x^2)(sin(1/x)
----------------------------------------------
h

How do I simplify from here?

Thanks in advance - I really want to develop a solid method to solving these.
 
Physics news on Phys.org
You're doing it too general know. That is, you're forming the quotient

[tex]\frac{f(x+h)-f(x)}{h}[/tex]

But here you know that x=0. So try form the quotient where x=0.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
5
Views
2K