Prove that $f(x)=x^4$ is continuous on $[0,\infty)$ I understand that we want $\varepsilon$ so that $|x^2+y^2||x+y||x-y|<\varepsilon$, and for $|x+y||x-y|<\varepsilon$ I can choose $\delta$ so that $\delta(2|y|+\delta)<\varepsilon$, but I don't know what to do with the $|x^2+y^2|$ part.
 A: Continuity is not going to be uniform. You want to show continuity at a fixed $y$, say. You will want $|x-y|<\delta$. Then $|x|<|y|+\delta$, so 
$$\tag{1}
|x^4-y^4|=|x^2+y^2|\,|x+y|\,|x-y|<(|y|+\delta)^2+y^2)\,(2|y|+\delta)\,|x-y|.
$$
To make this less that $\varepsilon$, you choose $\delta$ so that
$$
|x-y|<\frac\varepsilon{(|y|+\delta)^2+y^2)\,(2|y|+\delta)}.
$$
It is still not that obvious how to choose $\delta$. But if you first force $\delta<1$, say, then $(1)$ becomes 
$$
|x^4-y^4|=|x^2+y^2|\,|x+y|\,|x-y|<(|y|+\delta)^2+y^2)\,(2|y|+\delta)\,|x-y|
<((|y|+1)^2+y^2)(2|y|+1)\,|x-y|,
$$
and now you can choose 
$$
\delta=\min\left\{1,\frac\varepsilon{(|y|+1)^2+y^2)\,(2|y|+1)}\right\}.
$$
A: If $y=0$, then $|x^4-y^4|=x^4< \epsilon$ whenever $|x-y|=|x|<\epsilon^{1/4}$.  So, take $\delta=\epsilon^{1/4}$.
If $y\ne 0$, then take $|x-y|<y/2$ so that $y/2<x<3y/2$.
Then, we have
$$\begin{align}
|x^4-y^4|&=|x-y|(x+y)(x^2+y^2)\\\\
&<|x-y|\left(\frac52 y\right)\left(\frac{13}{4}y^2\right)\\\\
&=|x-y|\frac{65}{8}y^3\\\\
&<\epsilon
\end{align}$$
whenever $|x-y|<\frac{8}{65y^3}\epsilon$.  
So, let $\delta =\min\left(\frac{y}{2},\frac{8}{65y^3}\epsilon\right)$.  Then, for all $\epsilon>0$, $|x^4-y^4|<\epsilon$ for any $x>0$, 
