# Difinition of a derivative

1. Apr 14, 2005

### UrbanXrisis

let f be a function such that $$lim_{ h->x} \frac{f(2+h)-f(2)}{h} = 5$$

Which of the following must be true?

I. f is continuous at x=2.
II. F is differentiable at x=2.
III. The derivative of f is continuous at x=2.

I know (I) is true because it can be differentiated
I know that (II) is ture to because the derivative was found so it can be differentiable.
I dont know if III is true because the it doesnt tell me the limit of f'(x)

are these correct?

2. Apr 14, 2005

### what

That isn't the limit definition for a derivative unless x = 0, or maybe you typed it in wrong.

3. Apr 14, 2005

### dextercioby

It's okay...At least,for me...I edited your typo and advise u to use the code

\lim_{...} for the limit and \rightarrow for the "->"...

Daniel.

4. Apr 14, 2005

### Andrew Mason

I think this must be:

$$\lim_{h\rightarrow{0}} \frac{f(2+h)-f(2)}{h} = 5$$ (ie. 0, not x)

I agree, with one caveat: We don't know the domain of h. If h can be positive or negative, f would be differentiable at 2 and, therefore, continous.

True, subject to the above.
I think that if a function f(x) is differentiable at x, its derivative must be continuous at x.

AM

5. Apr 14, 2005

### hypermorphism

Not true. Counterexample: Let f(x) = x2sin(1/x) for x != 0, and f(x) = 0 when x = 0. Then f is differentiable at zero, but the derivative is not continuous at 0.

6. Apr 14, 2005

### UrbanXrisis

wait, you just said that x!=0, how can "f(x) = 0 when x = 0" when you just said x cant equal zero?

7. Apr 14, 2005

### hypermorphism

This is a piecewise function. f(x) has one rule when x is not zero (!=0), and another rule when x=0.

8. Apr 15, 2005

### HallsofIvy

Staff Emeritus
No, he didn't say that. He said that as long as x is NOT 0, you define
f(x) to be x2 sin(1/x) (which obviously can't be correct for x=0) but that if x IS 0, f(0)= 0.