<s>[INST] Precise definition of Multivariable limit I have encountered this limit: $\lim_{(x,y) to (0,0)} \frac{x^2+3y-7}{2+x^3-5y^2}$. We can substitute and find the limit to be $\frac{-7}{2}$since it is not a difficult limit, however I am trying to prove it rigorously and am getting stuck. I’m starting out with for all $\epsilon > 0$ there exists $\delta> 0$ such that $| \frac{x^2+3y-7}{2+x^3-5y^2}+\frac{7}{2}|<\epsilon $ whenever $\delta>\sqrt{x^2+y^2}$. I expanded the first absolute value and got $\frac{1}{2}|\frac{7x^3+2x^2-35y^2+6y}{x^3-5y^2+2}|$, but I cannot get past here to find any inequalities. I think it should be easy to do, since it’s a really easy limit there’s nothing undefined, but I’m stuck. Thank you. 
 [/INST] Note that
$$
\delta > \sqrt{x^2+y^2} \implies \delta > \lvert x\rvert,\, \delta>\lvert y\rvert
$$
Then $\delta > \sqrt{x^2+y^2}$ implies that
$$
\lvert 7x^3+2x^2-35y^2+6y \rvert < 7\delta^3 + 2\delta^2 +35\delta^2+6\delta = 7\delta^3 + 37\delta^2+6\delta
$$
Now, suppose without loss of generality that $\delta < 1/5$ (I picked this small number pretty arbitrarily, any small enough number works here) then
$$
\lvert x^3-5y^2+2 \rvert \geq 2 - \delta^3 - 5\delta^2 \geq 2-\frac{1}{5^3}-5\frac{1}{5^2} \geq 1
$$
Combining the inequalities we have (and noting that $\delta < 1/5$) we see that
$$
\frac{1}{2}\left\lvert \frac{7x^3+2x^2-35y^2+6y}{x^3-5y^2+2} \right\rvert < \frac{7\delta^3 + 37\delta^2+6\delta}{2} \leq \frac{7\delta + 37\delta+6\delta}{2} = 25\delta
$$
So, given $\epsilon >0$ picking $\delta = \min \{\epsilon /50, 1/5\}$ will give you the inequality
$$
\frac{1}{2}\left\lvert \frac{7x^3+2x^2-35y^2+6y}{x^3-5y^2+2} \right\rvert<\epsilon
$$
</s>