# Chapter 14: Partial Derivatives - Section 14.2 - Limits and Continuity in Higher Dimensions - Exercises 14.2 - Page 798: 64

Limit does not exist.

#### Work Step by Step

The polar-coordinates are defined as: $x= r \cos \theta , y = r \sin \theta$ and $r^2=x^2+y^2$ $\lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{(x,y) \to (0,0) } \dfrac{2x}{x^2+x+y^2}$ or, $=\lim\limits_{r \to 0} \dfrac{ 2 r \cos \theta}{r^2+r\cos \theta}$ or, $=\lim\limits_{r \to 0} \dfrac{ 2 \cos \theta}{r+\cos \theta}$ Suppose $a \cos \theta \to 0$; $\lim\limits_{a \cos \theta \to 0} \dfrac{ 2 \cos \theta}{ cos \theta(a+1) }=\dfrac{2}{a+1}$ Then we can see that for this case, the limit is not a unique value. Thus, $\lim\limits_{(x,y) \to (0,0) } f(x,y)=\lim\limits_{(x,y) \to (0,0) } \dfrac{2x}{x^2+x+y^2}$ =Limit does not exist.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.