Prove Continuous Function of t: R^2 to R

[ryo0071]

Okay so the question is:

Let \(\displaystyle f:R^2 \rightarrow R\) by
\(\displaystyle f(x) = \frac{x_1^2x_2}{x_1^4+x_2^2}\) for \(\displaystyle x \not= 0\)

Prove that for each \(\displaystyle x \in R\), \(\displaystyle f(tx)\) is a continuous function of \(\displaystyle t \in R\)

(\(\displaystyle R\) is the real numbers, I'm not sure how to get it to look right).

I am letting \(\displaystyle t_0 \in R\) and \(\displaystyle \epsilon > 0\) then trying to find a \(\displaystyle \delta > 0\) so \(\displaystyle |f(t) - f(t_0)| < \epsilon\) whenever \(\displaystyle |t - t_0| < \delta\) I am stuck trying to find the delta what will work, in trying to find it I am unable to simplify out \(\displaystyle |t - t_0|\) to use. Am I missing something really obvious here? Any help appreciated.

 

[Fantini]

Perhaps you mean that for each $x \in \mathbb{R}^2, x \neq 0 \in \mathbb{R}^2$ and $t \in \mathbb{R}$ the function $f(tx)$ is continuous? Because we have

$$f(tx) = f(tx_1, tx_2) = \frac{(tx_1)^2 (tx_2)}{(tx_1)^4 + (tx_2)^2} = \frac{t^3 x_1^2 x_2}{t^4 x_1^4 + t^2 x_2^2} = \frac{t^3 x_1^2 x_2}{t^2 (t^2 x_1^4 + x_2^2)} = \frac{tx_1^2 x_2}{t^2 x_1^4 + x_2^2}.$$

This function tends to zero as $t \to 0$ and is continuous everywhere else by noting that it is the result of operations with continuous functions (power, quotient, products and compositions).

EDIT: I think this needs a bit more explanation. If $x = (x_1, x_2) \neq 0$ then this means that $x_1 \neq 0$ or $x_2 \neq 0$ (this is a logical 'or', both can be nonzero). If $x_1 =0$ and $x_2 \neq 0$ then we obviously have $f(tx) = 0$ because the expression in the numerador is automatically zero while the denominator is nonzero. The same if the variables switch roles (the first becomes nonzero and the second becomes zero). Therefore the only case left to be discussed is when both are nonzero. Then you have what I just said. :)

 

[ryo0071]

Thank you for your response. I probably should have mentioned I have taken care of the cases where \(\displaystyle x_1 = 0\) and \(\displaystyle x_2 \not= 0\) as well as \(\displaystyle x_1 \not= 0\) and \(\displaystyle x_2 = 0\). Also, I am aware that it would be continuous since it is the result of operations of continuous function but I am trying to prove it using the epsilon-delta definition of the limit (by actually finding a delta that will work for an arbitrary epsilon, which is where I am getting stuck).

 

[Fantini]

I don't think you will manage to do it with the epsilon-delta definition. This function is the usual counterexample that you can have a function continuous at the origin for every line through it but it is actually discontinuous there: just consider the case where $x_2 = x_1^2$. In fact, you probably forgot to mention the definition at $x = 0$, else it is automatically continuous at where it is defined. If you manage to show this by epsilon-delta proof it would mean that it is continuous at the origin, which is not.

 

[blue_raver22]

To prove that f(tx) is a continuous function of t, we need to show that for any given x \in R, and for any \epsilon > 0, there exists a \delta > 0 such that for all t \in R, if |t - t_0| < \delta, then |f(tx) - f(t_0x)| < \epsilon.

To find the appropriate \delta, we can start by simplifying the expression |f(tx) - f(t_0x)|. Using the definition of f(x), we have:

|f(tx) - f(t_0x)| = \left|\frac{(tx_1)^2(tx_2)}{(tx_1)^4+(tx_2)^2} - \frac{(t_0x_1)^2(t_0x_2)}{(t_0x_1)^4+(t_0x_2)^2}\right|

= \left|\frac{t^2x_1^2x_2}{t^4x_1^4+t^2x_2^2} - \frac{t_0^2x_1^2x_2}{t_0^4x_1^4+t_0^2x_2^2}\right|

= \left|\frac{t^2x_1^2x_2}{t^4x_1^4+t^2x_2^2} - \frac{t_0^2x_1^2x_2}{t_0^4x_1^4+t_0^2x_2^2}\cdot\frac{t^4+t_0^4}{t^4+t_0^4}\right|

= \left|\frac{t^2x_1^2x_2}{(t^4+t_0^4)x_1^4+(t^2+t_0^2)x_2^2} - \frac{t_0^2x_1^2x_2}{(t^4+t_0^4)x_1^4+(t^2+t_0^2)x_2^2}\right|

= \left|\frac{(t^2-t_0^2)x_1^2x_2}{(t^4+t_0^4)x