Proving Continuity of a Function in R^2 Using Sequences

In summary, the function f is continuous at (0,0) if and only if the sequence a_n converging to (0,0) has an absolute value of less than r^2.
  • #1
jjou
64
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Let [tex]f:\mathbb{R}^2\rightarrow\mathbb{R}[/tex] be [tex]f(0,0)=0[/tex] and [tex]f(x,y)=\frac{x|y|}{\sqrt{x^2+y^2}}[/tex] for [tex](x,y)\neq(0,0)[/tex]. Is f continuous at (0,0)?



I tried showing it WAS NOT continuous by finding sequences that converge to 0 but whose image did not converge to 0. I tried sequences of the form (ct, t) where c was a constant and t went to 0 as well as sequences of the form (t^c, t). Simple forms such as (t^c, t^c) or (1/t, 1/t) did not work either.

Then I tried to show it WAS continuous by showing it was lipschitz, which turned into a horribly horribly long expansion without a clear inequality - so I'm pretty sure this isn't the correct method.

Is there a method I am overlooking?

(Also, am I allowed to ignore the absolute value in the numerator if I restrict (x,y) to the first and second quadrants of [tex]\mathbb{R}^2[/tex]?)
 
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  • #2
Brute force is always good. Take sqrt(x^2+y^2)=r. Then |x|<=r and |y|<=r. So the absolute value of the numerator is less than r^2. So |f(x,y)|<=r. Now let r->0.
 
  • #3
Thanks!

So, I define a sequence [tex](a_n)[/tex] such that, for each n, [tex]a_n=(x_n,y_n)\in\D_{1/n}(0,0)=\{a\in\mathbb{R}^2|d(a,0)=1\n\}[/tex]. Then as [tex]n\rightarrow\infty[/tex] we have [tex]a_n\rightarrow0[/tex]. Then for any n, [tex]\sqrt{x_n^2+y_n^2}=1/n[/tex] which implies that [tex]|x_n|\leq1/n[/tex] and [tex]|y_n|\leq1/n[/tex].

Then we have that [tex]f(a_n)\rightarrow f(0)=0[/tex] iff [tex]|f(a_n)|\rightarrow0[/tex].

[tex]|f(a_n)|=\left|\frac{x_n|y_n|}{\sqrt{x_n^2+y_n^2}}\right|=\frac{|x_n||y_n|}{\sqrt{x_n^2+y_n^2}}=n*|x_n||y_n|\leq n*\frac{1}{n}*\frac{1}{n} = \frac{1}{n}[/tex]

which completes the proof since [tex]\frac{1}{n}\rightarrow0[/tex]. Thus the function f is continuous.

Thanks again! :)
 
  • #4
Actually... I have to show this works for arbitrary sequence [tex]a_n[/tex] converging to (0,0). So I should define the sequence [tex]r_n=d(a_n,0)=\sqrt{x_n^2+y_n^2}[/tex] and the rest is the same.

:cool:
 

Related to Proving Continuity of a Function in R^2 Using Sequences

1. What is continuity of function in R^2?

Continuity of function in R^2 refers to the property of a function where there are no abrupt changes or breaks in its graph. This means that as the input values of the function approach a specific point, the output values also approach a specific point, without any sudden jumps or gaps.

2. How is continuity of function in R^2 different from continuity in R?

The main difference between continuity in R^2 and continuity in R is that continuity in R^2 takes into account two input variables, while continuity in R only considers one. This means that for a function to be continuous in R^2, it must satisfy the continuity condition in both the x and y directions.

3. What is the importance of continuity of function in R^2?

Continuity of function in R^2 is important because it allows us to make predictions and analyze the behavior of a function over a continuous range of values. It also helps us to identify any potential errors or inaccuracies in our calculations, as any abrupt changes or breaks in the function's graph can indicate a problem with the function itself.

4. Can a function be continuous in R^2 but not differentiable?

Yes, a function can be continuous in R^2 but not differentiable. A function is differentiable if it has a well-defined slope at every point, which means that the function must be smooth and have no sharp turns or corners. However, a function can be continuous in R^2 even if it has sharp turns or corners, as long as these points do not create any breaks or jumps in the graph.

5. How can we determine if a function is continuous in R^2?

A function is continuous in R^2 if it satisfies the three conditions of continuity: 1) the function is defined at the point in question, 2) the limit of the function as the input values approach the point exists, and 3) the limit is equal to the output value at that point. These conditions can be checked using calculus techniques such as the limit definition of continuity or the intermediate value theorem.

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