Simple Differentiability and Continuity Question

[Wormaldson]

Homework Statement

If f(x) = 3 for x < 0 and f(x) = 2x for x ≥ 0, is f(x) differentiable at x = 0? State and justify why/why not.

Homework Equations

The Attempt at a Solution

Obviously, since f(x) is not continuous and the limit doesn't exist as x[itex]\rightarrow[/itex]0, the function shouldn't be differentiable at that point. But I can't justify this because I don't really understand why not. Simply put, I get that the limit doesn't exist at 0 and, hence, the function is discontinuous there, but I don't see why that invalidates the idea that the gradient of the line f(x) = 2x at x = 0 is 2. Further, as far as I can tell (and I have a strong feeling I'm going wrong here somewhere), f'(x) = 0 for x < 0 and f'(x) = 2 for x ≥ 0, which seems to suggest that the derivative of f(x) for all values of x ≥ 0 is 2. What am I missing here?

As always, any help would be much appreciated, thanks.

 

[HallsofIvy]

You say, twice, that the limit does not exist at 0. What limit? I think you mean just [itex]lim_{x\to 0} f(x)[/itex] but the derivative, at x= 0, is defined as
[tex]\lim_{h\to 0}\frac{f(h)- f(0)}{h}= \lim{h\to 0} \frac{f(h)}{h}[/tex]
if the numerator does not go to 0, then that limit cannot exist.

 

[Wormaldson]

You say, twice, that the limit does not exist at 0. What limit? I think you mean just [itex]lim_{x\to 0} f(x)[/itex] but the derivative, at x= 0, is defined as
[tex]\lim_{h\to 0}\frac{f(h)- f(0)}{h}= \lim{h\to 0} \frac{f(h)}{h}[/tex]
if the numerator does not go to 0, then that limit cannot exist.

You're right; I was considering the problem in too narrow a scope and only considering the limit as x[itex]\rightarrow[/itex]0. So let's see: the limit as x[itex]\rightarrow[/itex]0 of f(x) does not exist, therefore, the limit as h[itex]\rightarrow[/itex]0 of f(0 + h) - f(0) does not exist either. The limit as h[itex]\rightarrow[/itex]0 of h is obviously just 0, and, since in [tex]\lim_{h\to 0}\frac{f(0 + h) - f(0)}{h}[/tex] the denominator approaches 0 as expected, but the numerator does not exist, then the limit of the quotient and hence the derivative does not exist. Is that right?

 

[HallsofIvy]

For this particular function, yes, the limit of the numerator does not exist. But the reason why a function has to be continuous to be differentiable is more general. Consider another example: f(x)= 2x for x not equal to 0, f(0)= 1. Now [itex]\lim_{h\to 0}f(h)= 0[/itex] but [itex]\lim_{h\to 0} f(h)- f(0)= -1[/itex]. The limit of the numerator does exist but is not 0 so [itex]\lim_{h\to 0} (f(h)- f(0))/h[/itex] does not exist.