Q 45. - Calculus

Question

In Exercises 41–46, use polar coordinates to analyze the given limits. 

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

Step-by-Step Solution

Answer

The value of $\underset{(x,y)\to \left(0,0\right)}{\lim } \frac{xy}{\sqrt{x^2+y^2}}$ is $0$.

Step 1. Given information.

We have given expression :  $\underset{(x,y)\to \left(0,0\right)}{\lim } \frac{xy}{\sqrt{x^2+y^2}}$

Step 2. Use polar coordinates to analyze the given limits.

The relation between the rectangular coordinates $(x,y)$ and the polar coordinates  $(r,\theta )$is

$$\begin{gathered} x=r \cos \theta \\ y=r \sin \theta \end{gathered}$$

On substituting values of  $x$ and $y$ we get.

$$\begin{gathered} \underset{(x,y)\to \left(0,0\right)}{\lim } \frac{xy}{\sqrt{x^2+y^2}}=\underset{r\to 0}{\lim } \frac{r \cos \theta r \sin \theta }{\sqrt{r^2 {\cos }^2\theta +r^2{\sin }^2\theta }} \\ =\underset{r\to 0}{\lim } \frac{r^2 \cos \theta \sin \theta }{r\sqrt{ {\cos }^2\theta +{\sin }^2\theta }} \\ =\underset{r\to 0}{\lim } r \cos \theta \sin \theta \\ =0 \end{gathered}$$

Since the value of $\underset{(x,y)\to \left(0,0\right)}{\lim } \frac{xy}{\sqrt{x^2+y^2}}$ is $0$.