Justifying implicit limits Consider the hyperbola given as:
$$\frac{x^2}{4}- \frac{y^2}{12} =1$$
Divide through by $x^2$
$$ \frac14 - \frac{1}{12} \left(\frac{y}{x}\right)^2= \frac{1}{x^2}$$
Now, here is the tricky step, I take limit as $ x \to \infty$ on both sides, this kills the RHS, and doing some algebra in the back, I get:
$$ 3= \lim_{x \to \infty} \left(\frac{y}{x}\right)^2$$
Now, I want to say that the slope of tangent line infinity/ asymptote to parabola has slope of $\sqrt{3}$ i.e: $ \lim_{x \to \infty} y' = 3$ and say that $ \lim_{x \to \infty} \left(y'\right)^2 = \left( \frac{y}{x}\right)^2$ is that possible?
I considered using Stolz Casero theorem for this, but for it I'd need a sequence, now how do I get a sequence such that I can say the above equivalence?

$\lim_{x\rightarrow \infty}\frac{y^2}{x^2}=3$ is a result you derived from the original equation. It contains relationship between x and y. We know that x is a set of constant values in R. In order for the equation to be true, y has to be the same order with x.
For example, if $y=x^2$ then $\lim_{x\rightarrow \infty}\frac{y^2}{x^2}=\lim_{x\rightarrow \infty}\frac{x^4}{x^2}=\lim_{x\rightarrow \infty}x^2=\infty$.
So it actually depends on your definition for y. If you use $y=f(x)=ax, a\in R$, you can solve the limit directly.\
If a function is continuous on $R$, for any point in it's co-domain, there must exist a sequence convergent to it. Since $f(x)$ is continuous, there must exist a sequence convergent to it.
