$$\frac{\pi x y^2}{4}$$

Is this function continuous? I really haven't worked with continuity with multivariable funtions before, so I am a little stumped. How would one answer such a question?

I'm reading a bit ahead of my level, and I'm seeing all these epsilon delta things... is that what I am supposed to use? Makes very little sense....

  • $\begingroup$ The function is $f(x,y)=\frac{\pi x y^2}{4}$? This is a continuous function. You don't need epsilon delta proofs for the intuition. The intuition of the function is that if you looked at the graph of the function, you don't see any jumps, holes, or any discrepancies like these (or infinite asymptotes in the middle of the graph!) But since there are none of these here, the function will look smooth. Can you imagine what it looks like? $\endgroup$ – Eoin Nov 1 '14 at 23:39
  • $\begingroup$ You can first prove $f(x,y) = x$ and $g(x,y) = y$ are both continuous. Then use the fact that the product of continuous functions are still continuous. $\endgroup$ – Petite Etincelle Nov 1 '14 at 23:40

Continuity in $\mathbb R^2$ of a function is defined as follows:

Let $X$ be an open subset of $\mathbb R^2$ and $f$ a real valued function from $X \subseteq \mathbb R^2 \rightarrow \mathbb R$. We say that $f$ is continuous over $X$ if $\forall (x_0,y_0)\in X\wedge\forall\varepsilon>0\;\exists\,\delta >0$ such that:

$$\sqrt{(x_0-x)^2+(y_0-y)^2}<\delta \Rightarrow |f(x,y)-f(x_0,y_0)|<\varepsilon.$$

Is the function $f(x,y)=\dfrac{\pi}{4}$ continuous?

Let $\delta \in \mathbb R^+$, then $\sqrt{(x_0-x)^2+(y_0-y)^2}<\delta \Rightarrow |f(x,y)-f(x_0,y_0)| = 0<\varepsilon$.

Is the function $f(x,y)=x$ continuous?

$|x-x_0|=|x_0-x|=\sqrt{(x_0-x)^2}\leq\sqrt{(x_0-x)^2+(y_0-y)^2}<\delta=\varepsilon$. From which it can be seen that if $\sqrt{(x_0-x)^2+(y_0-y)^2}<\delta=\epsilon$ then $|f(x,y)-f(x_0,y_0)|<\varepsilon$.

Do the same with $y$. Recall that the product of continuous functions is continuous as well.


You are asking if the function $f(x,y) = \frac{\pi xy^{2}}{4}$ is continuous? If you had a functions of x or y alone, then it'd be easy to see that it's continuous, right? Together, they should still be continuous... I hope this agrees with your intuition. Proving that the function is continuous, you just need the definition of continuity and the definition of the limit for multivariate functions. Check this out!



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