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?
 A: Hint: To finish ...
$$0 \leqslant\frac{(\sqrt{x^{2}+y^{2}})|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}\leqslant \sqrt{x^{2}+y^{2}}\left(\frac{|x^{2}+y^{2}|}{|x^{2}+y^{2}|}+\frac{|xy|}{|x^{2}+y^{2}|}\right) \leqslant \sqrt{x^{2}+y^{2}}(\ldots)$$
A: 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.
A: You can use polar coordinates:
$$x=r\cos t; y=r \sin t;\\
\lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=\lim_{r\to 0}\frac{r^{2}(1+\frac12\sin 2t)}{|r|}=0.$$
A: You may also procced as follows using GM-QM (inequality between geometric and quadratic mean):


*

*$\sqrt{ab}\leq \sqrt{\frac{a^2+b^2}{2}}$
So, you get 
\begin{eqnarray*}  \left| \frac{x^2+xy+y^2}{\sqrt{x^2+y^2}}\right|
 & = & \left| \frac{x^2+y^2}{\sqrt{x^2+y^2}} + \frac{xy}{\sqrt{x^2+y^2}}\right| \\
 & \leq  &   \sqrt{x^2+y^2} + \frac{|xy|}{\sqrt{x^2+y^2}}\\
 & \stackrel{GM-QM}{\leq} & \sqrt{x^2+y^2} + \frac{\left(\sqrt{\frac{x^2+y^2}{2}} \right)^2}{\sqrt{x^2+y^2}} \\
& = & \frac{3}{2}\sqrt{x^2+y^2} \\
& \stackrel{(x,y)\to (0,0)}{\longrightarrow} & 0
\end{eqnarray*}
