Limit Exercise for \(f(x,y)\)

Exercise:

Compute, $\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,\frac{4x{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}$

Solution: we know that ${{y}^{2}}\le {{x}^{2}}+{{y}^{2}}\,\,\And \,\,\,{{x}^{2}}+{{y}^{2}}\ge 0$

$\Leftrightarrow \frac{1}{{{x}^{2}}+{{y}^{2}}}\le \frac{1}{{{y}^{2}}}\Leftrightarrow \frac{4x{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\le 4x$ but $\frac{4{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\ge 0\,\,\,\And \,\,x\le x{{y}^{2}}$

Hence $0\le \frac{4x{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\le 4x$ as $\left( x,y \right)\to 0$

By squeeze theorem $\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,\frac{4x{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}=0$

2nd method by using $x=r\cos \theta \,\,\And \,\,y=r\sin \theta $ so ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$

As $x\to 0\,,\,r\to 0$

So \[\underset{\left( x,y \right)\to \left( 0,0 \right)}{\mathop{\lim }}\,\frac{4x{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}=\underset{r\to 0}{\mathop{\lim }}\,\frac{4r\cos \theta {{r}^{2}}{{\sin }^{2}}\theta }{{{r}^{2}}}=\underset{r\to 0}{\mathop{\lim }}\,2r\sin 2\theta \sin \theta =0\]