Without relying at the graph of the following function;$$y^2=\frac{x^4}{x^2-1}$$How does one evaluate the following limits: $\lim\limits_{x \to ∞}y$ and $\lim\limits_{x \to -∞}y$ ?

I am clueless as to where to start.

  • 2
    $\begingroup$ Supossing the second limit the inverse of the first, you can assert that when $x\to \pm\infty$ the limiting value of thefunction $y$ clearly $y\to\pm\infty$ since $y^2\sim x^4/x^2=x^2$. Contrary when $y\to\pm \infty$ the variable $x$ must do the same $i.e$ $x^4\sim y^2$ and therefore $x\to\pm\infty$ $\endgroup$
    – HBR
    Aug 1, 2017 at 11:30
  • $\begingroup$ Apologies, slight mistake in the question wording - without relying, not without looking. Graph may be viewed but not used in the explanation. $\endgroup$ Aug 1, 2017 at 11:37
  • $\begingroup$ Did I explain myself correctly in the previous comment? $\endgroup$
    – HBR
    Aug 1, 2017 at 12:58
  • $\begingroup$ @HBR you certainly did, I'm just referring back to the answer given in the question which seems to contradict the logical answer given here. $\endgroup$ Aug 4, 2017 at 2:51

1 Answer 1


For $|x|>1$ we have $y(x)= \pm \frac{x^2}{\sqrt{x^2-1}}$.

For example: it holds $\frac{x^2}{\sqrt{x^2-1}} \ge x$ if $x>1$, hence

$\frac{x^2}{\sqrt{x^2-1}} \to \infty$ for $x \to \infty$.

It is your turn to consider the other cases.


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