Let D = {(x, y) ∈ R
2
: x > 0, 0 < y < x3}. Define
f(x, y) = (
0, (x, y) ∈/ D
1, (x, y) ∈ D.
a) Approaching (0, 0) along the line y = mx for each real number m and the y-axis, prove that
lim(x,y)→(0,0) f(x, y) exists and compute the limit.
b) Argue whether f is continuous at (0, 0).
(a)
lim(x,y)→(0,0)f(x,y)=lim(x,y)→(0,0)f(x,mx)So,lim(x,y)→(0,0)f(x,0)≤lim(x,y)→(0,0)f(x,y)≤lim(x,y)→(0,0)(mx)3⇒0≤lim(x,y)→(0,0)f(x,y)≤0Hence,lim(x,y)→(0,0)f(x,y)=0\lim\limits_{(x,y)\rightarrow (0,0)} f(x,y)\\ =\lim\limits_{(x,y)\rightarrow (0,0)} f(x,mx)\\ So, \lim\limits_{(x,y)\rightarrow (0,0)} f(x,0)\leq\lim\limits_{(x,y)\rightarrow (0,0)} f(x,y)\leq\lim\limits_{(x,y)\rightarrow (0,0)} (mx)^3\\ \Rightarrow 0\leq\lim\limits_{(x,y)\rightarrow (0,0)} f(x,y)\leq0\\ Hence, \lim\limits_{(x,y)\rightarrow (0,0)} f(x,y)=0(x,y)→(0,0)limf(x,y)=(x,y)→(0,0)limf(x,mx)So,(x,y)→(0,0)limf(x,0)≤(x,y)→(0,0)limf(x,y)≤(x,y)→(0,0)lim(mx)3⇒0≤(x,y)→(0,0)limf(x,y)≤0Hence,(x,y)→(0,0)limf(x,y)=0
(b) Yes, the function is continuous at x=0.x=0.x=0.
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