how-to-prove-this-function-is-not-continuous-at-the-origin.

[aruwin]

Please show me the solution for this math problem. I have no idea how to do this. (This is not homework. I am studying for my exam).

Show that the following function is not continuous at the origin(0,0).

 

[daon2]

For the function to be continuous it must be continuous at every method of approach. Try approaching (0,0) along the "path" y=sqrt(x). Then try approaching (0,0) along the x axis. Is the limit the same?

 

[aruwin]

For the function to be continuous it must be continuous at every method of approach. Try approaching (0,0) along the "path" y=sqrt(x). Then try approaching (0,0) along the x axis. Is the limit the same?

How to approach the path?

 

[pka]

How to approach the path?

For examples:

Path I: $\displaystyle x=y$

Path II: $\displaystyle x=y^2$

 

[aruwin]

For examples:

Path I: $\displaystyle x=y$

Path II: $\displaystyle x=y^2$

So I have to find the limits for these,right? And then what do I do?

 

[pka]

So I have to find the limits for these,right? And then what do I do?

If the two limits are different, then the function is not continuous.

 

[aruwin]

If the two limits are different, then the function is not continuous.

Can you teach me how to find the limits?

 

[pka]

On path I: $\displaystyle \displaystyle\lim _{y \to 0} \dfrac{{y^3 }}{{y^2 + y^4 }}=~?$

 

[aruwin]

On path I: $\displaystyle \displaystyle\lim _{y \to 0} \dfrac{{y^3 }}{{y^2 + y^4 }}=~?$

Well,yeah. Actually, I need help with all the paths.

 

[aruwin]

Oh,I get it now!!I can choose any path that I want as long as it shows that the limit is not equals to 0,right? So I would just take one path 2 because it's easier to calculate. The limit would be a constant, 1/2. Thus this shows that it is not continous with f(0,0). So then it is proven! Am I right?Or not?

 

[pka]

Oh,I get it now!!I can choose any path that I want as long as it shows that the limit is not equals to 0,right? So I would just take one path 2 because it's easier to calculate. The limit would be a constant, 1/2. Thus this shows that it is not continous with f(0,0). So then it is proven! Am I right?Or not?

In this case what path you choose must pass through $\displaystyle (0,0)$.
On path I the limit is $\displaystyle 0~.$

On path II the limit is $\displaystyle \frac{1}{2}$.

So not continuous.