<s>[INST] Solve the following quadratic equation: $4\tan^2(\theta)x^4-4x^2+1=0$ $$4\tan^2(\theta)x^4-4x^2+1=0$$
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?
 [/INST] $$\lim_{\theta\to0}\frac{1-\sqrt{1-\tan^2\theta}}{2\tan^2\theta}
= \lim_{\theta\to0}\frac{(1-\sqrt{1-\tan^2\theta})(1+\sqrt{1-\tan^2\theta})}{(2\tan^2\theta)(1+\sqrt{1-\tan^2\theta})} = \lim_{\theta\to0}\frac{\tan^2\theta}{(2\tan^2\theta)(1+\sqrt{1-\tan^2\theta})}=\lim_{\theta\to0}\frac{1}{2(1+\sqrt{1-\tan^2\theta})}=\frac{1}{2}$$
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