Question

1. By considering different paths of approach, show that the function has no limit as (x, y) -------> (0, 0).

Answer 1

Answer:

1. shown below

2. $\frac{-1}{\sqrt{2}}$

Step-by-step explanation:

We say $\displaystyle \lim_{(x,y)\rightarrow (a,b)}f(x,y)$ exists if the limit remains the same along every path.

Here, f is a function on two variables defined on a disk that contains the point (a,b).

1.

Along y-axis i.e., x = 0:

$\displaystyle \lim_{(x,y)\rightarrow (a,b)}f(x,y)=\displaystyle \lim_{(x,y)\rightarrow (0,0)}\frac{-x}{\sqrt{x^2+y^2}}=\displaystyle \lim_{y\rightarrow 0}\frac{0}{\sqrt{0^2+y^2}}=0$

Along x-axis i.e., y = 0:

$\displaystyle \lim_{(x,y)\rightarrow (a,b)}f(x,y)=\displaystyle \lim_{(x,y)\rightarrow (0,0)}\frac{-x}{\sqrt{x^2+y^2}}=\displaystyle \lim_{x\rightarrow 0}\frac{-x}{\sqrt{x^2+0^2}}=\displaystyle \lim_{x\rightarrow 0}\frac{-x}{x}=-1$

As the limit is not the same along different paths, so limit does not exist.

2.

Along the path x = y:

$\displaystyle \lim_{(x,y)\rightarrow (a,b)}f(x,y)=\displaystyle \lim_{(x,y)\rightarrow (0,0)}\frac{-x}{\sqrt{x^2+y^2}}=\displaystyle \lim_{x\rightarrow 0}\frac{-x}{\sqrt{x^2+x^2}}=\displaystyle \lim_{x\rightarrow 0}\frac{-x}{\sqrt{2}x}=\frac{-1}{\sqrt{2}}$