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 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. $