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Here's the limit, it's so crazy i've no idea how to evaluate it, but I know it equals $0$. $$\lim_{u,v\to 0,0}\frac{|u^2+u^2v+v^2u+2vu|}{\sqrt{u^2+v^2}}.$$

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    $\begingroup$ Are you sure you mean $u^2+u^2$ in the numerator? $\endgroup$ – Git Gud May 27 '14 at 21:50
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Let $f(u,v) = \dfrac{\left|u^2+u^2+v^2u+2vu\right|}{\sqrt{u^2+v^2}}$.

Clearly, $f(u,v) \geq 0$, so the limit (should it exist) is non-negative. Now, note that for $|u|,|v|<1$, we have

$$\begin{align} |u^2+u^2+v^2u+2vu| &\leq |u|^2 + |u|^2 + |v^2| |u| + 2 |v| |u|\\ &\leq |u|^2 + |u|^2 + |v|^2 + 2 |u||v|\\ &= |u|^2 + (|u|+|v|)^2\\ &\leq |u|^2 + 2 (|u|^2+|v|^2)\\ &\leq |u|^2+|v|^2+ 2 (|u|^2+|v|^2)\\ &= 3(|u|^2+|v|^2)\\ &= g(u,v),\end{align}$$ by triangle inequality, $|u|<1 \implies |v^2||u| \leq |v|^2$, and the fact that $(a+b)^2 \leq 2 (a^2+b^2)$ for $a,b>0$, and $|v|^2\geq 0$.

Now, since the numerator and denominator of $f(u,v)$ are non-negative, we see that $f(u,v) \leq \dfrac{g(u,v)}{\sqrt{u^2+v^2}} = \dfrac{3 \left(|u|^2+|v|^2\right)}{\sqrt{u^2+v^2}} = 3 \sqrt{u^2+v^2}$.

Now, you know that $\sqrt{u^2+v^2} \to 0$ as $(u,v) \to (0,0)$,
so $\lim \limits_{(u,v) \to (0,0)} f(u,v) \leq \lim \limits_{(u,v) \to (0,0)} 3 \sqrt{u^2+v^2} = 0$.

As stated prior, we have that this limit is $\geq 0$ since $f(u,v)$ is non-negative. Thus, it must be $0$.

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