Continuity of Multivariable Functions

[Chewy0087]

EDIT: just realized i might've been really stupid;

very simple question which will answer my stupidly long question;

is f(x) = 1 continuous?]

The reason I ask is that my book says;

f(x,y) \in C^{N} in R \Leftrightarrow \frac{\partial ^{n} f}{\partial x^n} , \frac{\partial ^{n} f}{\partial x^{n-1}\partial y}, etc \in C in R.

however you can imagine for f (x, y) = x which IS continuous,

f1(x, y) = 1 , would this be considered continuous, and if not, isn't that at ends with the definition above?

 

[Dick]

Of course it's continuous. lim (x1,y1)->(x,y) f(x1,y1)=1. f(x,y)=1. It's easy using any definition of continuity. Look one up!

 

[Chewy0087]

Yes, I thought as much, but why then would;

f(x) = x (for x greater than/equal to 0) / 0 (for x less than 0)

Why then for this function would

<br /> f(x) \notin C^{1}<br />

?

Would it be because there is a jump, a "discontinuity" at 0, where f(x) goes from 1 to 0 instantly?

 

[Dick]

Yes, I thought as much, but why then would;

f(x) = x (for x greater than/equal to 0) / 0 (for x less than 0)

Why then for this function would

<br /> f(x) \notin C^{1}<br />

?

Would it be because there is a jump, a "discontinuity" at 0, where f(x) goes from 1 to 0 instantly?

Sure. f(x) is continuous. f'(x) is not continuous (I'm assuming you meant to say f'(x)). So it's not in C^1.

 

[Chewy0087]

Hmmm so it's not continuous?

But then sureley it is at ends with the definition posted earlier; namingly

<br /> f(x,y) \in C^{N}<br /> <br /> \Leftrightarrow \frac{\partial ^{n} f}{\partial x^n} , \frac{\partial ^{n} f}{\partial x^{n-1}\partial y}, etc \in C<br />

because df/dx is not continuous, so it is implied that f(x) is not continuous, or is there a flaw in my reasoning?

edit: and yeah sorry i meant f'(x)

 

[Chewy0087]

sorry I just realized I'm being a major douche...i'm stupid, nevermind