# Continuity Question

• Oct 20th 2013, 06:38 PM
Assassin0071
Continuity Question
Okay so the question is:

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

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

([MTEXR[/MTEX is the real numbers, I'm not sure how to get it to look right).

I am letting $t_0 \in R$ and $\epsilon > 0$ then trying to find a $\delta > 0$ so $|f(t) - f(t_0)| < \epsilon$ whenever $|t - t_0| < \delta$ I am stuck trying to find the delta what will work. I start with $|f(t) - f(t_0)|$ and try and simplify but I am not sure how to get to the point where I can determine delta. Any help appreciated.
• Oct 20th 2013, 07:24 PM
SlipEternal
Re: Continuity Question
I think what you want is: \mathbb{R}

Anyway, let's calculate $f(tx)$. Note that $f(0x)$ is undefined, so $f(tx)$ can only be a continuous function of $t \in \mathbb{R}\setminus \{0\}$.

$f(tx) = \dfrac{(tx_1)^2(tx_2)}{(tx_1)^4 + (tx_2)^2} = \dfrac{tx_1^2x_2}{t^2x_1^4 + x_2^2} = \dfrac{tx_1^{-2}x_2}{t^2+(x_1^{-2}x_2)^2}$

Note: The final simplification is only possible if $x_1\neq 0$. If $x_1=0$, then $f(tx)=0$ is a constant function and obviously continuous. Similarly, if $x_2=0$, you also have a constant function, which is obviously continuous. So, assume $x_1\neq 0$ and $x_2\neq 0$.

Since $f(-tx) = -f(tx)$, the function is odd. So, you only need to consider continuity for $t>0$ and by symmetry, you will find continuity for all $t\in \mathbb{R}\setminus \{0\}$.

Now, given $x\in \mathbb{R}^2 \setminus \{(0,0)\}$, $t_0\in \mathbb{R}\setminus \{0\}$, and $\varepsilon>0$, you want to find $\delta>0$ that works. First of all, we are going to want to limit the possible range of $\delta$. We know that the function is not continuous at $t=0$, so $0 < \delta < |t_0|$.

Does this help at all?