Respuesta :
Answer:
1. shown below
2. [tex]\frac{-1}{\sqrt{2}}[/tex]
Step-by-step explanation:
We say [tex]\displaystyle \lim_{(x,y)\rightarrow (a,b)}f(x,y)[/tex] 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:
[tex]\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[/tex]
Along x-axis i.e., y = 0:
[tex]\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[/tex]
As the limit is not the same along different paths, so limit does not exist.
2.
Along the path x = y:
[tex]\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}}[/tex]
The value of limit along two different path is not same. Therefore, no limit exist.
The value of limit along y=x is [tex]-\frac{1}{\sqrt{2} }[/tex]
limit of function:
The given function is,
[tex]f(x,y)=\frac{-x}{\sqrt{x^{2} +y^{2} } }[/tex]
Considering first path is x- axis i.e. y=0
[tex]\lim_{(x,y) \to (0,0)} f(x,y)= \lim_{(x,y) \to (0,0)} \frac{-x}{\sqrt{x^{2} +0^{2} } }=-1[/tex]
Considering another path is y-axis i.e. x=0
[tex]\lim_{(x,y) \to (0,0)} f(x,y)= \lim_{(x,y) \to (0,0)} \frac{0}{\sqrt{0^{2} +y^{2} } }=0[/tex]
Since, the value of limit along two different path is not same. Therefore, no limit exist.
Value of limit along path y = x.
[tex]\lim_{(x,y) \to (0,0)} f(x,y)= \lim_{(x,y) \to (0,0)} \frac{-x}{\sqrt{x^{2} +y^{2} } }=-\frac{1}{\sqrt{2} }[/tex]
Learn more about the limit of function here:
https://brainly.com/question/23935467
