Proving Continuity of a Function Using Epsilon-Delta Definition

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Homework Statement



Hey guys, I've been given the following \epsilon - \delta proof question. The trouble I'm having is I'm not 100% sure what it is actually asking for, and how to go about it. The more common proofs involving two variable functions and a given limit are easy enough to do, but I'm stuck on this one.

Homework Equations



Suppose f: \left( \mathbb R ^2 , \| . \|_2 \right) \rightarrow \left( \mathbb R , | .| \right) is continuous at \left(a,b \right) \in \mathbb R^2. Prove using the \epsilon - \delta definition only, that if we define the function

f_b : \left( \mathbb R,|.| \right) \rightarrow \left( \mathbb R,|.| \right) s.t. f_b \left(x \right) = f \left( x,b \right),

then f_b is continuous at x=a.

The Attempt at a Solution



Well, considering I am not actually sure what the question is acting for I can't go very far. I was assuming you would set up the usual proof as,

Let \epsilon > 0
if |(x,b) - (a,b)| < \delta
then |f(x,b) - (a,b)| < \epsilon

Possibly then, x - a < \delta?

Any help would be appreciated, cheers.
 
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Polter19 said:

Homework Statement



Hey guys, I've been given the following \epsilon - \delta proof question. The trouble I'm having is I'm not 100% sure what it is actually asking for, and how to go about it. The more common proofs involving two variable functions and a given limit are easy enough to do, but I'm stuck on this one.

Homework Equations



Suppose f: \left( \mathbb R ^2 , \| . \|_2 \right) \rightarrow \left( \mathbb R , | .| \right) is continuous at \left(a,b \right) \in \mathbb R^2. Prove using the \epsilon - \delta definition only, that if we define the function

f_b : \left( \mathbb R,|.| \right) \rightarrow \left( \mathbb R,|.| \right) s.t. f_b \left(x \right) = f \left( x,b \right),

then f_b is continuous at x=a.

The Attempt at a Solution



Well, considering I am not actually sure what the question is acting for I can't go very far. I was assuming you would set up the usual proof as,

Let \epsilon > 0
if |(x,b) - (a,b)| < \delta
then |f(x,b) - (a,b)| < \epsilon

Possibly then, x - a < \delta?

Any help would be appreciated, cheers.

Take \varepsilon >0 fixed.

You know that f is continuous at (a,b), thus you know that there exists a \delta>0 such that for all (x,y) it holds that

\|(x,y)-(a,b)\|_2<\delta~\Rightarrow~|f(x,y)-f(a,b)|<\varepsilon

Now, what you must do is to show f_b continuous at a. Thus you must find a \delta>0 such that for all x it holds that

|x-a|<\delta~\Rightarrow~|f(x,b)-f(a,b)|<\varepsilon

Can you proceed now?
 
I am also interested in this question. I understand what is required for the proof, but I am just stuck as to how to manipulate:
|f(x,b) - f(a,b)|
into something that resembles δ??

We know that:
0 < ||(x,y) - (a,b) || < δ \Rightarrow |f(x,y) - f(a,b)| < ε
as the limit is equal to the value of the function at (a,b) since it is continuous there, but how can we follow on from this to show that if y is restricted to equalling b, then:
0 < ||(x,b) - (a,b) || = ||x-a|| < δ \Rightarrow |f(x,b) - f(a,b)| < ε
?? I'm really stuck and any help would be appreciated.
 
jj22vw25 said:
I am also interested in this question. I understand what is required for the proof, but I am just stuck as to how to manipulate:
|f(x,b) - f(a,b)|
into something that resembles δ??

We know that:
0 < ||(x,y) - (a,b) || < δ \Rightarrow |f(x,y) - f(a,b)| < ε
as the limit is equal to the value of the function at (a,b) since it is continuous there, but how can we follow on from this to show that if y is restricted to equalling b, then:
0 < ||(x,b) - (a,b) || = ||x-a|| < δ \Rightarrow |f(x,b) - f(a,b)| < ε
?? I'm really stuck and any help would be appreciated.

How is ||(x, y)- (a, b)|| defined? (there are several equivalent definitions- which are you using?)
 
||(x,y) - (a,b)|| is a norm measuring the distance from (a,b) to (x,y), what do you mean how is it defined? Distance between points in R^2, euclidean norm.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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