Q 41. - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1. Given information.

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

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.

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

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