Limit of two variable Suppose that I have equation:
$$\tan(a) = \dfrac{b-c}{bc + 1},\;\text{ where }\,a, b, c\, \text{ are variables.}$$
How can I show that, as $a \to \pi/2, \; bc+1\to 0\;$ mathematically, not intuitively.
I know how to calculate limit, as variable $(X,Y)$ approaches to $(X_0,Y_0)$, but not the other way. 
How can I find point where $(X,Y)$ approaches, when the limit is given, mathematically rather than intuitively?
 A: If $A \to \pi/2$ then $\tan(A)\to +\infty$ supposed the value of $A$ is ever less than $\pi/2$, because $\tan$ is continuous on the interval $(-\pi/2,\pi/2)$. Since our equality holds, $bc+1=\frac{c^2+1}{1-c\tan a}$, so there are $b$ and $c$ for every $a$ that $bc+1=1$, which shows we need another assumptions too to show $bc+1\to 0$.
A: We know that as $a$ goes to $\pi/2$, $\tan(a)$ goes to infinity.  So in the right hand side of your expression either the numerator goes to infinity or the denominator goes to zero.  If the numerator goes to infinity, either $b$ goes to infinity or $c$ goes to -infinity (or both).  But you can see if b or c have this behavior, the fraction does not go to infinity.  Therefore, the only alternative is that $bc+1$ goes to zero.
A: Solving for $c$, we have
$$ c = \frac{ b - \tan a }{ b \tan a + 1 } = \frac{b \cos a- \sin a}{b \sin a + \cos a}$$
Then, writing $b^\star$ and $c^\star$ for $\lim b$ and $\lim c$ as $a\to\pi/2$ (assuming both limits exist), 
$$c^\star = \frac{b^\star \cos\frac{\pi}{2}-\sin\frac{\pi}{2}}{b^\star \sin\frac{\pi}{2}+\cos\frac{\pi}{2}} = \frac{-1}{b^\star}$$
so that $b^\star c^\star + 1 = 0$. That is, $bc+1\to 0$, as claimed. $\square$
