# Finding limit of multivariable function using the squeeze theorem

In my textbook (Stewart's Calculus), the video tutor solutions for some problems use the squeeze theorem to determine the limit of a function. For example:

Find $$\lim_{(x, y) \to (0, 0)} \frac{x^2y^3}{2x^2+y^2}.$$

The typical solution I keep seeing involves taking the absolute value of $f(x, y)$ and then using some properties of inequalities to deduce the limit using the squeeze theorem, like so:

$0 \leq |\frac{x^2y^3}{2x^2+y^2}| \leq |y^3|$ because $x^2 \leq 2x^2 + y^2$ and thus $\frac{x^2}{2x^2+y^2} \leq 1$

Then, taking the limit of both sides of the inequality yields $L = 0$.

But that's the limit for the absolute value of $f(x, y)$, not $f(x, y)$. Why is this approach valid?

• In any normed space, if $|x|\to 0$ then $x\to 0$ – user160738 Jan 31 '17 at 15:05
• @user160738 Ohh....that simple, huh? – AleksandrH Jan 31 '17 at 15:08
• @SimplyBeautifulArt Not quite: $f(x,y)$ could be negative, for instance. – Clement C. Jan 31 '17 at 15:10
• @ClementC. Oopsies, my bad. :3 – Simply Beautiful Art Jan 31 '17 at 15:10
• By definition, in a normed space$X$, $x\to 0$ means $\left \| x \right \|\to 0$, because the norm induces the (norm) topology on $X$ – Matematleta Jan 31 '17 at 15:13

In short: because $f(\vec{u}) \xrightarrow[\lvert \vec{u}\rvert \to 0]{} 0$ if, and only if, $\lvert f(\vec{u})\rvert \xrightarrow[\lvert \vec{u}\rvert \to 0]{} 0$.
You can check it by the $\varepsilon$/$\delta$ definition of the limit, for instance: $f(\vec{u})$ converges to $L$ when $\vec{u}\to a$ means that, for every $\varepsilon>0$, there exists $\delta>0$ such that $\lvert \vec{u} - a\rvert \leq \delta$ implies $\lvert f(\vec{u}) - L\rvert \leq \varepsilon.$