$$4\tan^2(\theta)x^4-4x^2+1=0$$
It can be seen that as $\theta$ goes to $0$, the leftmost term disappears and $x=\pm \frac{1}{2}$.
But when I try to solve this using the quadratic formula I get:
$$x=\pm \sqrt{\frac{1\pm \sqrt{1-\tan^2(\theta)}}{2\tan^2(\theta)}}$$
In this expression as $\theta$ goes to $0$ the denominator goes to zero.
I guess there may be a way of manipulating this expression such that as $\theta$ goes to zero $x$ goes to $\pm 0.5$.
How could this expression be manipulated?
On
$$\lim_{\theta\to0}\frac{1-\sqrt{1-\tan^2\theta}}{2\tan^2\theta} = \lim_{\theta\to0}\frac{(1-\sqrt{1-\tan^2\theta})(1+\sqrt{1-\tan^2\theta})}{(2\tan^2\theta)(1+\sqrt{1-\tan^2\theta})} = \lim_{\theta\to0}\frac{\tan^2\theta}{(2\tan^2\theta)(1+\sqrt{1-\tan^2\theta})}=\lim_{\theta\to0}\frac{1}{2(1+\sqrt{1-\tan^2\theta})}=\frac{1}{2}$$
Basically, your question can be reformulated to
To calculate that, first of all, since $\sqrt{}$ is a continuous function, all you need to do is to caluclate $$\lim_{\theta\to 0}\frac{1\pm \sqrt{1-\tan^2(\theta)}}{2\tan^2(\theta)}$$ and take the square root.
This limit has two distinct possibilities:
The first:
Taking the plus sign makes $$\lim_{\theta\to 0}\frac{1+ \sqrt{1-\tan^2(\theta)}}{2\tan^2(\theta)} = \infty$$
which makes sense because, as $\theta$ becomes small but not zero, two of the roots of your polynomial become larger and larger.
The second:
We can introduce a new variable $y=\tan(\theta)$ and get
$$\lim_{\theta\to 0}\frac{1- \sqrt{1-\tan^2(\theta)}}{2\tan^2(\theta)} = \lim_{y\to 0} \frac{1-\sqrt{1-y^2}}{2y^2}$$
which can easily be calculated either by using L'Hospital or by seeing that $$\frac{1-\sqrt{1-y^2}}{2y^2} = \frac{1-\sqrt{1-y^2}}{2y^2}\cdot\frac{1+\sqrt{1-y^2}}{1+\sqrt{1-y^2}} = \frac{1-(1-y^2)}{2y^2(1+\sqrt{1-y^2})} =\\=\frac{y^2}{2y^2(1+\sqrt{1-y^2})} = \frac{1}{2(1+\sqrt{1-y^2})}$$