Precise definition of Multivariable limit

Question

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.

Answer 1

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 $$ \frac{1}{2}\left\lvert \frac{7x^3+2x^2-35y^2+6y}{x^3-5y^2+2} \right\rvert<\epsilon $$

Answer 2

For a function $f: \mathbb{R^n} \to \mathbb{R}$ the definition is the same as for that for a function of one variable but in this case we need to check the $\epsilon$ condition $|f(x,y)-L|<\epsilon$ for

$$\forall (x,y) \in B_{\delta}(x_0)\setminus \{x_0\}$$

that is the point $(x,y)$ belongs to the open circle (ball) centered in $x_0$ with $x_0$ excluded.

Therefore for $f: \mathbb{R} \to \mathbb{R}$ the definition is

$$\left(\lim_{x\rightarrow a} f(x) = L\right) \iff \Big(\forall \varepsilon >0\, \exists \delta > 0: \big(0<\vert x-a\vert <\delta \implies \vert f(x)-L\vert <\varepsilon\big)\Big)$$

and for $f: \mathbb{R^n} \to \mathbb{R}$ the definition is

$$\left(\lim_{\vec x\rightarrow \vec a} f(\vec x) = L\right) \iff \Big(\forall \varepsilon >0\, \exists \delta > 0: \forall \vec x \in B_{\delta}(\vec x_0)\setminus \{\vec x_0\} \implies \vert f(\vec x)-L\vert <\varepsilon\big)\Big)$$

Note that analogous definitions holds for the cases with $L=\pm \infty$ and $|\vec a|=\infty$.