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How to find range of $2$ variable function $\displaystyle g(x,y)=\frac{\sqrt{5-x}}{y^2}$ for real $x,y$

My way

First i calculate domain of that function

Here $5-x\geq 0\;\cap\; y\neq 0\Longrightarrow x\leq 5\;\cap\; y\in\mathbb{R}-\{0\}$

And for calculation of range

Largest value of $g(x,y)\rightarrow \infty$ when $y\rightarrow \pm \infty$ but how i find lowest value of range

But i did not understand how to find range of function

Please help me to find range of function

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    $\begingroup$ Are $x$ and $y$ real numbers? What is the domain of $g$? Is it the largest subset of the real numbers for which $g(x,y)$ is again a real number? Are you familiar with complex numbers? Also, there are a few speling mistakes and mathematical errors; presumably you mean $\wedge$ instead of $\cap$. $\endgroup$
    – Servaes
    Mar 30, 2020 at 19:40
  • $\begingroup$ Largest value of $g(x,y)\rightarrow \infty$ when $y\rightarrow \pm \infty$ but how i find lowest value of $g(x,y)$ $\endgroup$
    – jacky
    Mar 30, 2020 at 19:46
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    $\begingroup$ If $x,y\in\mathbb R$, then the numerator is non-negative and the denominator is positive; the lowest value of $g(x,y)$ occurs when $x=5$; also, when $y\to\pm\infty,$ $g(x,y)\to0$ $\endgroup$
    – J. W. Tanner
    Mar 30, 2020 at 19:48
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    $\begingroup$ @jacky: That's what I mean, so the range of $g(x,y)$ is contained in $[0,\infty)$; can you find $x$ and $y$ such that $g(x,y)=a$ for any $a\in[0,\infty)$? $\endgroup$
    – J. W. Tanner
    Mar 30, 2020 at 19:53
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    $\begingroup$ Lowest value is exactly $0$ when $x=5$ $\endgroup$
    – jacky
    Mar 30, 2020 at 19:54

1 Answer 1

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Suppose $g$ has real inputs and outputs only. Then consider the domain: $x \leq 5$ and $y \neq 0$. The numerator can be as small as $0$ and as large as you'd like ($x=-105$ yeilds a numerator of 10). The denominator can be "very close" to $0$ and as large as you like. Notice that you cannot get negative results. So, the smallest the output can be is $0$ ($x=5, y\neq 0$). How big can the output get? As long as $x$ is not 5, we can make the number very big, for simplicity let $x=4$, so that the numerator takes value 1. Now, $1/y^2$ can be made as large as you like. Say you want to make it size M, choose $y = 1/\sqrt{M}$. In other words, for all $x <5$, $$\lim_{y\to 0}g(x,y) =\infty $$ So $g \in [0,\infty)$.

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