Q 43. - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1. Given information.

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

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

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

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

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

$$\begin{gathered} \underset{(x,y)\to \left(0,0\right)}{\lim } \frac{x^2{y}^2}{x^2+y^2}=\underset{r\to \theta }{\lim } \frac{r^2{\cos }^2\theta \left(r^2{\sin }^2\theta \right)}{r^2{\cos }^2\theta +r^2{\sin }^2\theta } \\ =\underset{r\to \theta }{\lim } \frac{r^4{\cos }^2\theta {\sin }^2\theta }{r^2\left({\cos }^2\theta +{\sin }^2\theta \right)} \\ =\underset{r\to \theta }{\lim } r^2{\cos }^2\theta {\sin }^2\theta \\ =0 \end{gathered}$$

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