Prove that $\lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=0$.

Question

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?

Answer 1

Hint: To finish ...

$$0 \leqslant\frac{(\sqrt{x^{2}+y^{2}})|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}\leqslant \sqrt{x^{2}+y^{2}}\left(\frac{|x^{2}+y^{2}|}{|x^{2}+y^{2}|}+\frac{|xy|}{|x^{2}+y^{2}|}\right) \leqslant \sqrt{x^{2}+y^{2}}(\ldots)$$

Answer 2

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.

Answer 3

You can use polar coordinates: $$x=r\cos t; y=r \sin t;\\ \lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=\lim_{r\to 0}\frac{r^{2}(1+\frac12\sin 2t)}{|r|}=0.$$

Answer 4

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*}