Note: The function is poorly written, so we'll assume a better expression. You can use this answer as a guide for the real question
Answer:
Explanation below
Step-by-step explanation:
Two-Variable Limits
Given a two-variable function f(x,y), we need to compute if
[tex]\lim\limits_{(x,y) \rightarrow (0,0)}f(x,y)[/tex]
exists, being
[tex]\displaystyle f(x,y)=\frac{x^2y}{x^4+y^2}[/tex]
a) We'll use the function y=x to approach the point (0,0):
[tex]\displaystyle \lim\limits_{(x,y) \rightarrow (0,0)}\frac{x^2y}{x^4+y^2}[/tex]
Since y=x
[tex]=\displaystyle \lim\limits_{(x,y) \rightarrow (0,0)}\frac{x^3}{x^4+x^2}[/tex]
Factoring
[tex]=\displaystyle \lim\limits_{(x,y) \rightarrow (0,0)}\frac{x^3}{x^2(x^2+1)}[/tex]
Simplifying
[tex]=\displaystyle \lim\limits_{(x,y) \rightarrow (0,0)}\frac{x}{x^2+1}=0[/tex]
The limit gives 0
(b) Now we'll use [tex]y=x^2[/tex] to approach the point (0,0). We compute the limit again
[tex]\displaystyle \lim\limits_{(x,y) \rightarrow (0,0)}\frac{x^2y}{x^4+y^2}[/tex]
Plugging in [tex]y=x^2[/tex]
[tex]\displaystyle =\lim\limits_{(x,y) \rightarrow (0,0)}\frac{x^4}{x^4+x^4}[/tex]
Simplifying
[tex]\displaystyle =\lim\limits_{(x,y) \rightarrow (0,0)}\frac{1}{2}[/tex]
The limit gives 1/2
Since both limits are different, the limit does not exist
