Q 42. - Calculus

Question

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

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

Step-by-Step Solution

Answer

The $\underset{(x,y)\to \left(0,0\right)}{\lim } \frac{y^2}{x^2+y^2}$ does not exist.

Step 1. Given inforantion.

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

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.

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

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