Question

I want answers qualily​

Answer 1

Step-by-step explanation:

lim

(x, y) =(0,0) 2xy/rootx²+y²

we can see if we the zero in this equation u will get 0/0 form(indeterminant form) OK friend

lim

(x, y) =(0,0)

$\frac{2xy}{ \sqrt{ {x}^{2} + {y}^{2} } } \times \frac{ \sqrt{ {x}^{2} + {y}^{2} } }{ \sqrt{ {x}^{2} + {y}^{2} } }$

$\frac{2xy \sqrt{ {x}^{2} + {y}^{2} } }{ {x}^{2} + {y}^{2} }$

$\frac{2xy \sqrt{ {x}^{2} + {y}^{2} } }{( {x + y)}^{2} }$

$2xy$

$2(0)(0) = 0$

OK friend

u can also solve By Left hand limit and right hand limit u will get answer 0

hence LHL=RHL the limit exist. at point x=0