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Evaluate the limits :
a) lim (x,y)-->(0,0) (x^2 + y^2)/(√(x^2 +y^2 +1-1))
b) lim (x,y)-->(0,0) (3x^2 y^2)/(x^2 + y^2)
5 years ago

Answers

oobleck
I assume you mean
(x^2 + y^2)/(√(x^2 +y^2 +1)-1))
Let u^2 = x^2+y^2+1
Then you have
(u^2-1)/(u-1) = u+1 for u≠1
So, the limit is 2

(3x^2 y^2)/(x^2 + y^2) = 3/(1/y^2 + 1/x^2)
so the limit is 0
or, maybe easier to see,
(3x^2 y^2)/(x^2 + y^2) = 3x^2/(1+y^2/x^2)
5 years ago

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