I am trying to calculate the following:
$$\lim_{||x||\to\infty} \sup_{0 \leq t \leq 2\pi} \cfrac{||f(t,x)||}{||x||^7}$$
where $x = (x_1,x_2,x_3)$ and $f(t,x)= ((x_2^2+x_3^2+1)\cos(t),(x_1^2+x_3^2)\sin(t),(x_1^2+x_2^2)\cos(t))$
My attempt:
Try spherical coordinates, so $||x||^2=r^2$ and so $||x||^7= r^7$ and $$||f(t,x)||= \sqrt{(r^2-r^2\cos^2(\theta)\sin^2(\phi))^2\cos^2(t)+(r^2-r^2\cos^2(\phi))^2\cos^2(t)+(r^2-r^2\sin^2(\phi)\sin^2(\theta))^2\sin^2(t)}$$
We can factor out a $r$ out of the radical above, so we have (inside the supremum):
$\cfrac{||f(t,x)||}{||x||^7}=\cfrac{\sqrt{(1-\cos^2(\theta)\sin^2(\phi))^2\cos^2(t)+(1-\cos^2(\phi))^2\cos^2(t)+(1-\sin^2(\phi)\sin^2(\theta))^2\sin^2(t)}}{r^6}$
The supremum of $\cos(t)$ occurs at $t=0$ and supremum of $\sin(t)$ occurs at $t=\cfrac{\pi}{2}$.
So we can get that the supremum is $\leq\cfrac{\sqrt{(1-\cos^2(\theta)\sin^2(\phi))^2+(1-\cos^2(\phi))^2+(1-\sin^2(\phi)\sin^2(\theta))^2}}{r^6}$
Thus, letting $r \rightarrow \infty$ we get that the limit in question is $\leq 0$. I want to conclude that the limit is indeed $0$, which seems right enough. But I'm not sure if this is correct. Can someone please check what I did?
The numerator is proportional to $r^2$, not $r$, so the denominator after cancellation is $r^5$.
The numerator after cancellation is a continuous function on the compact unit sphere, and thus has an upper bound. Thus the supremum is bounded by a constant times $r^{-5}$, and thus goes to zero.
You don’t really need the whole spherical coordinate thing – the numerator goes as $r^2$ and the denominator goes as $r^7$, so the limit for $r\to\infty$ is $0$.