<s>[INST] Prove that $\lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=0$. 
Prove that $\displaystyle \lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=0$.

This has been my rough work so far, but I am not sure how to go further or if I am doing it the wrong way...
$\displaystyle|f(x,y)-L|=\left |\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}  \right | \leq \frac{|x^{2}+xy+y^{2}|}{\big|\sqrt{x^{2}+y^{2}}\big|}\cdot \frac{\sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}}} = \frac{\big(\sqrt{x^{2}+y^{2}}\big)|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}$
I wanted the square root on top to help determine what I should set my $\delta$ as. Any ideas on how to finish this or do it in a better way?
 [/INST] To begin with, notice that
\begin{align*}
\begin{cases}
|x| = \sqrt{x^{2}} \leq \sqrt{x^{2}+y^{2}}\\\\
|y| = \sqrt{y^{2}} \leq \sqrt{x^{2}+y^{2}}
\end{cases}\Longrightarrow
\begin{cases}
\displaystyle\frac{|x|}{\sqrt{x^{2}+y^{2}}} \leq 1\\\\
\displaystyle\frac{|y|}{\sqrt{x^{2}+y^{2}}} \leq 1
\end{cases}
\end{align*}
Furthermore, the given expression can be split as the following sum
\begin{align*}
E(x,y) = \frac{x^{2}+2xy+y^{2}}{\sqrt{x^{2}+y^{2}}} = \frac{x^{2}}{\sqrt{x^{2}+y^{2}}} + \frac{2xy}{\sqrt{x^{2}+y^{2}}} + \frac{y^{2}}{\sqrt{x^{2} + y^{2}}}
\end{align*}
As a consequence of the squeeze theorem applied to each summand, the given limit tends to zero.
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