How can I find the range of $f(x,y) = \frac{\sqrt{x-2y^2+1}}{x-2-y}$?

I first determined restrictions on the domain.

Namely, $x-2y^2+1 \ge 0$ (must have square root of positive number) rearranged to get:

$$y^2 \le\frac{x+1}{2}$$

Also $x-2-y \ne 0$ (denominator ) rearranged to get:

$$y \ne x-2$$

Here is the domain graphed: Graph of Domain (I do not have 10 reputation yet, so I cannot embed images)

I can see that at point $(0,0)$, the range can be negative ($f(0,0) = -\frac{1}{2}$)

I also see that when I let $y=0, \lim_{x\to\infty}{f(x)} = 0$ (not much help - it does not tend to positive nor negative infinity)

EDIT: I just realized that when I let $y=0, \lim_{x\to0^+}{f(x)} = \infty$
(therefore the range must include $0 \le f(x) \le \infty$)

Now, how can I find the range in the negatives?


Let denominator $=x-y-2 = \epsilon$
Then $x=y+2+\epsilon$
so $f(x,y)= \frac{\sqrt {(y-2y^2+3)+\epsilon)}}{\epsilon}$

If you vary $\epsilon$ from a very small positive number to a very small negative number, but not including $\epsilon =0$, and also because you can control $y-2y^2+3$ to be anything from $0$ to $3$ for eg., you will get every value from $-\infty$ to $\infty$

  • $\begingroup$ But I have the two restrictions (I can see how to get from 0 to infinity, but not 0 to negative infinity) $\endgroup$ – peqhusus Sep 23 '17 at 5:31
  • $\begingroup$ @peqhusus $\epsilon = -h; h \rightarrow 0$ implies $f(x,y) \rightarrow -\infty$. To get values closer to zero, minimise numerator by varying $y$, as $y-2y^2+3$ can be anything from $3$ to $0$ $\endgroup$ – john doe Sep 23 '17 at 5:36
  • $\begingroup$ @peqhusus and your restrictions are taken care of as I am not taking $\epsilon = 0$, and I won't consider values of $y$ for which numerator is negative. $\endgroup$ – john doe Sep 23 '17 at 5:39
  • $\begingroup$ I'm not familiar with the epsilon-delta notation. Is there any way to write it without? $\endgroup$ – peqhusus Sep 23 '17 at 6:01
  • $\begingroup$ @peqhusus I'm not using epsilon-delta, all I'm saying is that by varying the denominator $\epsilon$ (or don't call it epsilon if it bothers you, call it $t$) I can get every value from $-\infty$ to $\infty$. This is not a rigorous proof, it's kinda intuitive $\endgroup$ – john doe Sep 23 '17 at 6:06

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