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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?

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    In any normed space, if $|x|\to 0$ then $x\to 0$ – 2017-01-31
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    @user160738 Ohh....that simple, huh? – 2017-01-31
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    @SimplyBeautifulArt Not quite: $f(x,y)$ could be negative, for instance. – 2017-01-31
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    @ClementC. Oopsies, my bad. :3 – 2017-01-31
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    By definition, in a normed space$X$, $x\to 0$ means $\left \| x \right \|\to 0$, because the norm induces the (norm) topology on $X$ – 2017-01-31
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    And if I get 0 on the left hand side and, say, 1 on the right hand side, does that imply the limit does not exist? – 2017-01-31
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    No, it just means your application of the squeeze theorem did not succeed. It could be because the limit does not exist, or because your upper bound is too loose. – 2017-01-31
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    But take for example: $\frac{y^4}{x^4 + 6y^4}$. Since this function is positive and always less than or equal to 1, couldn't I use the squeeze theorem to conclude that the limit DNE, since the left side yields 0 and the right side of the inequality yields 1? – 2017-01-31
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    No, the squeeze theorem allows you to prove convergence when it works; when it does not, it doesn't prove divergence. To show the limit does not exist, you should instead find two different sequences $(x_n,y_n)_n$ and $(x'_n,y'_n)_n$ both converging to zero, but with different limits for $f(x_n,y_n)$ and $f(x'_n,y'_n)$, for instance. – 2017-01-31
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    E.g., here choose $(x_n,x_n)_n$: $f(x_n,y_n) = \frac{1}{7} \to \frac{1}{7}$. Now, choose $(x_n,0)_n$: $f(x_n,y_n) = 0 \to 0$. – 2017-01-31

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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. $