<s>[INST] Help evaluating $\lim_{\left|x\right| \to \infty}y$, given $\frac{y^2}{x^2}=\frac{b^2}{a^2}-\frac{b^2}{x^2}$ I'm trying to understand this step in a derivation of the standard equation of a hyperbola. We have constants $a$ and $b$, and variables $x$ and $y$. We've gotten to a point where we have
$$\frac{y^2}{x^2}=\frac{b^2}{a^2}-\frac{b^2}{x^2}$$
The text then states:

As x and y attain very large values, the quantity $\frac{b^2}{x^2}\to0$, so that $\frac{y^2}{x^2}\to \frac{b^2}{a^2}$. This shows that $y\to\pm\frac{b}{a}x$ as $\left|x\right|$ grows large.

I'm having a hard time getting from the first sentence to the second. How do we make that logical jump?
 [/INST] Since $$\left(\frac yx\right)^2=\frac{y^2}{x^2}\to\frac{b^2}{a^2}=\left(\frac ba\right)^2$$ as $|x|$ grows large, then $$\left(\frac yx-\frac ba\right)\left(\frac yx+\frac ba\right)=\left(\frac yx\right)^2-\left(\frac ba\right)^2\to 0$$ as $|x|$ grows large, so $$\frac yx-\frac ba\to 0\quad\text{or}\quad\frac yx+\frac ba\to 0$$ as $|x|$ grows large, so $$\frac yx\to\frac ba\quad\text{or}\quad\frac yx\to-\frac ba$$ as $|x|$ grows large, so $$\frac yx\to\pm\frac ba$$ as $|x|$ grows large. Then $y\sim\pm\frac bax$ as $|x|$ grows large.
We can't actually take the limit of $y$ as $|x|\to\infty$, as $|y|$ necessarily grows without bound.

An alternate way to see this, if you're still having trouble, is to note that $$y^2=\frac{b^2}{a^2}x^2-b^2,$$ so $$\left(y-\frac bax\right)\left(y+\frac bax\right)=y^2-\left(\frac bax\right)^2=y^2-\frac{b^2}{a^2}x^2=-b^2.$$ Thus, $$\lim_{x\to\infty}\left(y-\frac bax\right)\left(y+\frac bax\right)=-b^2\in\Bbb R,\tag{$\heartsuit$}$$ but since $\left|\frac bax\right|\to\infty$ as $x\to\infty$, then $(\heartsuit)$ can only happen if $y-\frac bax\to 0$ or $y+\frac bax\to 0$ as $x\to\infty$, so $y\to\pm\frac bax$ as $x\to\infty$. Similarly, $y\to\pm\frac bax$ as $x\to-\infty$, so $y\to\pm\frac bax$ as $|x|\to\infty$.
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