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\begin{center}
\textbf{Partial Differentiation}

\textit{\textbf{Limits}}
\end{center}

\textbf{Question}

If $\ds \lim_{(x,y)\to(0,0)} \frac{x^my^n}{(x^2+y^2)^p}$ is to exist,
what condition must the integers $m$, $n$ and $p$ satisfy, given that
$m,n,p>0$.

Prove your answer.


\textbf{Answer}

As
\begin{eqnarray*}
|x| & \le & \sqrt{x^2+y^2} \\
\textrm{and } |y| & \le & \sqrt{x^2+y^2}
\end{eqnarray*}
This gives
\begin{eqnarray*}
\left | \frac{x^my^n}{(x^2+y^2)^p} \right | & \le &
\frac{(x^2+y^2)^{(m+n)/2}}{(x^2+y^2)^p} \\
& & = (x^2+y^2)^{-p+(m+n)/2}\\
\textrm{And }(x^2+y^2)^{-p+(m+n)/2} & \to & 0 \textrm{ as }(x,y)\to(0,0)\\
\textrm{provided } m+n & > & 2p
\end{eqnarray*}

$$\Rightarrow \lim_{(x,y)\to (0,0)} \frac{x^my^n}{(x^2+y^2)^p}=0.$$

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