Limit exercise for a function of several variable x ,y

Exercise:

Compute, $\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,\frac{xy\left( {{x}^{2}}-{{y}^{2}} \right)}{{{x}^{2}}+{{y}^{2}}}$

Solution: Let $y=\lambda x$ as $x\to 0\,,y\to 0$ where $\lambda \in \mathbb{R}$

So $\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,\frac{xy\left( {{x}^{2}}-{{y}^{2}} \right)}{{{x}^{2}}+{{y}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\lambda {{x}^{2}}\left( {{x}^{2}}-\lambda {{x}^{2}} \right)}{{{x}^{2}}+\lambda {{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\lambda {{x}^{2}}\left( {{x}^{2}}\left( 1-\lambda  \right) \right)}{{{x}^{2}}\left( 1+\lambda  \right)}$
$=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}\lambda \left( 1-\lambda  \right)}{1+\lambda }=0$

Or: by using polar coordinates as follows :

Take $x=r\cos \theta \,\,\And \,\,\,y=r\sin \theta \,\,\Leftrightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}$

So \(\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,\frac{xy\left( {{x}^{2}}-{{y}^{2}} \right)}{{{x}^{2}}+{{y}^{2}}}=\underset{r\to 0}{\mathop{\lim }}\,\frac{{{r}^{2}}\cos \theta \sin \theta \left( {{r}^{2}}\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta  \right) \right)}{{{r}^{2}}}\)

                                \(=\frac{1}{2}\underset{r\to 0}{\mathop{\lim }}\,{{r}^{2}}\sin 2\theta \cos 2\theta =0\)

( $\theta $ is undefined at zero )

Or: using inequalities

We know that  ${{x}^{2}}+{{y}^{2}}+2xy\ge 0\Leftrightarrow {{x}^{2}}+{{y}^{2}}\ge 2xy$ and ${{x}^{2}}+{{y}^{2}}>0$

So $0<\frac{1}{{{x}^{2}}+{{y}^{2}}}\le \frac{1}{2xy}\Leftrightarrow 0<\frac{xy}{{{x}^{2}}+{{y}^{2}}}\le \frac{1}{2}\Leftrightarrow 0<\frac{xy\left( {{x}^{2}}-{{y}^{2}} \right)}{{{x}^{2}}+{{y}^{2}}}\le \frac{1}{2}\left( {{x}^{2}}-{{y}^{2}} \right)$

as $\left( x,y \right)\to \left( 0,0 \right)$ so ${{x}^{2}}-{{y}^{2}}\to 0$ hence by squeeze theorem $\frac{xy\left( {{x}^{2}}-{{y}^{2}} \right)}{{{x}^{2}}+{{y}^{2}}}\to 0$





Exercise:

Show that the limit of $f\left( x,y \right)=\frac{x{{y}^{3}}}{{{x}^{2}}+{{y}^{6}}}$ doesn’t exists at $\left( 0,0 \right)$

Solution: Let $y=\lambda x\,\,,\,\,\lambda \in \mathbb{R}$ so $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\lambda }^{3}}{{x}^{4}}}{{{x}^{2}}+{{\lambda }^{6}}{{x}^{6}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\frac{1}{{{\lambda }^{3}}{{x}^{2}}}+{{\lambda }^{3}}{{x}^{2}}}=\frac{1}{\infty }=0$

So there exists a limit through this path $y=\lambda x$

But $\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,\frac{x{{y}^{3}}}{{{x}^{2}}+{{y}^{6}}}=\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,\frac{\frac{x{{y}^{3}}}{x{{y}^{3}}}}{\frac{{{x}^{2}}}{x{{y}^{3}}}+\frac{{{y}^{6}}}{x{{y}^{3}}}}=\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,\frac{1}{\frac{x}{{{y}^{3}}}+\frac{{{y}^{3}}}{x}}$

Now let’s consider new path $u=\frac{x}{{{y}^{3}}}$ with $y=\lambda \sqrt[3]{x}\,\,\,,\lambda \in {{\mathbb{R}}^{*}}$

So $\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,\frac{x}{{{y}^{3}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{{{\lambda }^{3}}x}=\frac{1}{{{\lambda }^{3}}}$ hence $\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,u\left( x,y \right)=\frac{1}{{{\lambda }^{3}}}$

So $\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,\frac{1}{\frac{x}{{{y}^{3}}}+\frac{{{y}^{3}}}{x}}=\underset{u\left( x,y \right)\to \frac{1}{{{\lambda }^{3}}}}{\mathop{\lim }}\,\frac{1}{u\left( x,y \right)+\frac{1}{u\left( x,y \right)}}=\frac{1}{\frac{1}{{{\lambda }^{3}}}+{{\lambda }^{3}}}$

Put $\lambda =1$ to get $\underset{u\to 1}{\mathop{\lim }}\,\frac{1}{\frac{1}{u}+u}=\frac{1}{1+1}=\frac{1}{2}$ so in the same path we get different values

depend on the value of $\lambda $ hence the there is a limit on this path but not unique

Thus the limit doesn’t exists at zero