Q 46. - Calculus

Question

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

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

Step-by-Step Solution

Answer

The value of $\underset{(x,y)\to \left(0,0\right)}{\lim } \frac{\sin (x^2+y^2)}{x^2+y^2}$is $1$.

Step 1. Given information.

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

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.

$$\begin{gathered} \underset{(x,y)\to \left(0,0\right)}{\lim } \frac{\sin (x^2+y^2)}{x^2+y^2}=\underset{r\to 0}{\lim } \frac{\sin (r^2{\cos }^2\theta +r^2{\sin }^2\theta )}{r^2{\cos }^2\theta +r^{2 }{\sin }^2\theta } \\ =\underset{r\to 0}{\lim } \frac{\sin r^2 \left({\cos }^2\theta +{\sin }^2\theta \right)}{r^2\left({\cos }^2\theta +{\sin }^2\theta \right)} \\ =\underset{r\to 0}{\lim } \frac{\sin r^2 }{r^2} \\ =\underset{r\to 0}{\lim }\frac{2r \cos \left(r^2\right) }{2r} \\ =\underset{r\to 0}{\lim } \cos \left(r^2\right) \\ =\cos 0 \\ =1 \end{gathered}$$

Since the value of $\underset{(x,y)\to \left(0,0\right)}{\lim } \frac{\sin (x^2+y^2)}{x^2+y^2}$ is $1$.