Extreme values for a vector equation For a question on physics.stackexchange about Does the Ampère-Maxwell law fail for the field of a uniformly moving point charge? with
$$ \vec B(P) = \dfrac{\mu_0 q}{4 \pi} \dfrac{1 - v^2/c^2}{[1 - (v^2/c^2) sin^2 \phi]^{3/2}} \dfrac{\vec v \times \hat r}{r^2} $$
for this sketch

I want to know the extreme value for $v \rightarrow c$ and have doubts about my thoughts:
A possible case is $\phi =0°$. For this angle the equation gives
$$ \vec B(P) = \dfrac{\mu_0 q}{4 \pi} \dfrac{1 - v^2/c^2}{1} \dfrac{\vec v \times \hat r}{r^2} $$
But what is about the term $\dfrac{\vec v \times \hat r}{r^2}$? It is a vector product and for 0° it is zero and $ \vec B(P)$ equals zero?
Taking $\phi =90°$ I’m lost because I don’t know what is the solution for  $v \rightarrow c$ of 
$$ \dfrac{1 - v^2/c^2}{[1 - (v^2/c^2)]^{3/2}} $$
 A: From the picture we have $0\le y=r\sin(\pi-\phi)$, so $\sin\phi=-\frac yr$. Put $A=\frac{\mu_0qc^2}{4\pi}$, $\vec u=\frac {\vec v}{c}$. I assume that $\mu_0\ne 0$, $r\ne 0$, and $\hat r=\frac {\vec r}{r}$.  Then
$$\vec B(P)=A\cdot\frac{1-u^2}{\left(1-u^2\frac {y^2}{r^2}\right)^{3/2}}\cdot \frac{\vec u\times\vec r}{r^3}.$$
Assume first that the charge needs to go a long way to be accelerated to $c$, so $r$ tends to infinity. Then   $\vec B(P)$ tend to zero because $A$ is constant, $ u^2\frac {y^2}{r^2}$ tends to zero, $1-u^2$ is bounded, and $|\vec u\times\vec r |=O(r)=o(r^3)$.
As an opposite case from now I assume that $\vec r$ is constant. 
Assume first that $x\ne 0$. Then $y<r$, so the denominator $1-u^2\frac {y^2}{r^2}\ne 0$ and $B(\vec P)$ is a continuous function of $u$. Therefore $\lim_{u\to 1} \vec B(P)=\vec B(P)|_{u=1}=0$, because when $u=1$ we have $1-u^2=0$. 
Assume now that $x=0$. Then $y=r$. Since $\vec u=(u,0,0)$ and $\vec r=(0,r,0)$, we have  
$$\frac{\vec u\times\vec r}{r^3}=\frac 1{r^3} \cdot \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
u & 0 & 0 \\
0 &  r  & 0 
\end{vmatrix}=\frac{u\mathbf{k}}{r^2}.$$
For $v=c$ the value of $\vec B(P)$ is undefined, because the denominator $\left(1-u^2\frac {y^2}{r^2}\right)^{3/2}=0$. For $v<c$ we have 
$$\vec B(P)=A\cdot\frac{1-u^2}{\left(1-u^2\right)^{3/2}}\cdot \frac{\vec u\times\vec r}{r^3}= \frac{Au\mathbf{k}}{\left(1-u^2\right)^{1/2}r^2}.$$
$$\lim_{u\to 1, u\ne 1} \vec B(P)= \lim_{u\to 1, u\ne 1} \frac{Au\mathbf{k}}{\left(1-u^2\right)^{1/2}r^2}=\infty.$$
Now about the growth of $\vec B(P)$ when $v$ tends to $c$, that is $u$ tends to $1$. We have  
$$|\vec B(P)|=\left|A\cdot\frac{1-u^2}{\left(1-u^2\frac {y^2}{r^2}\right)^{3/2}}\cdot \frac{\vec u\times\vec r}{r^3}\right|=\frac{C(1-u^2)u}{\left(1-a u^2\right)^{3/2}},$$ where $C=\left|A\cdot \frac{(1,0,0)\times\vec r}{r^3}\right|$ and $a=\frac {y^2}{r^2}\le 1$. 
If  $x\ne 0$ then $0\le a<1$, and the derivative $\left(\frac{(1-u^2)u}{\left(1-a u^2\right)^{3/2}}\right)'$ is $\frac{1-(3-2a)u^2}{(1-au^2)^{5/2}}$, so  $|\vec B(P)|$  monotonically grows when $u$ grows from $0$ to $\frac{1}{\sqrt{3-2a}}$ and then monotonically decreases to $0$ when $u$ grows to $1$. 
If $x=0$ then $a=1$  and $|\vec B(P)|=\frac{Cu}{\left(1-u^2\right)^{1/2}}$. This value monotonically grows to the infinity when $u$ monotonically tends to $1$. 
