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Homework 8

  1. (30 points)
    Show if the following sets are domains or not:

    1. \(A = \{(x,y) \in \mathbb{R}^{2}: x^{2}+ y^{2}> 1\}\).

      Yes

      Since \(A = \{(x,y) \in \mathbb{R}^{2} : x^{2}+y^{2} > 1\} = \mathbb{R}^{2} \setminus \overline {B(0,1)}\) and \(\overline{B(0,1)}\) is closed.
      Then \(A\) is open.

      For any points \(p=(x_1,y_1)\) and \(q=(x_2,y_2)\), there are three segments can ensure that they are connected

      1. Segment \(pq\)​
      2. If segment \(pq\) cannot connect them on \(A\), then consider two segment: 1. segment from \(p=(x_1,y_1)\) to \((x_1,y_2)\) 2. segment from \((x_1,y_2)\) to \(q=(x_2,y_2)\)​
      3. If it doesn't work, then consider three segment: 1. segment from \(p=(x_1,y_1)\) to \((1.5,y_{1})\) 2. segment from \((1.5,y_1)\) to \((1.5,y_2)\) 3. segment from \((1.5,y_2)\) to \(q=(x_2,y_2)\)​

      Then we can also find a path

      image​

      Thus it's path connected, then it's a domain 2. \(B = \{(x,y) \in \mathbb{R}^2 : xy > 0\}\)

      No

      NTP: ① \(B = B^{\circ}\) ② connected
      ① Since \(B^{\circ}\subseteq B\) is trivial by definition. Then we just prove \(B \subseteq B^{\circ}\).
      Take \((x,y) \in B\). where \(xy>0\). NTP: \(\exists r>0 : B((x,y),r) \subseteq B\). Let \(r = \min\{\frac{|x|}{2}, \frac{|y|}{2}\}\).

      I) \(x>y>0\). \(r=\frac{y}{2}\). Take \((a,b) \in B((x,y),r)\). \((a,b) = (x+t \cos \theta, y+t \sin \theta)\) where \(t=kr,0<k<1\).
      ​\(x+t\cos\theta\ge x+\frac{yk}{2}\cos\theta\geq x-\frac{yk}{2}\geq\frac{2y-yk}{2}> 0\).
      ​\(y+t\sin\theta\ge y+\frac{yk}{2}\cos\theta\geq y-\frac{yk}{2}\geq\frac{2y-yk}{2}> 0\). Thus \(ab>0\).
      II) \(y>x>0\). \(r = \frac{x}{2}\). Similarly.
      III) \(x<y<0\). \(r = -\frac{y}{2}\). Similarly.
      IV) \(y<x<0\). \(r = -\frac{x}{2}\). Similarly. Thus \(B=B^{\circ}\).

      ② Since \(B = \{(x,y) : x>0, y>0\} \cup \{(x,y) : x<0, y<0\}\).
      Let \(A = \{(x,y) : x>0, y>0\}\) and \(C = \{(x,y) : x<0, y<0\}\).
      A, C are open. Since \(A \cup C = B\) is open. Then B is disconnected.
      Thus B is not domain. 3. \(C = \{(x,y) \in \mathbb{R}^{2}: y < x^{2}\}\).

      Yes

      NTP ① \(C=C^{\circ}\). ② C is connected.

      ① Let \(f(x,y) = x^2 - y\).
      Since \(C = \{(x,y) \in \mathbb{R}^2 : y < x^2\} = \{(x,y) \in \mathbb{R}^2 : x^2 - y > 0\}\).
      So, \(C = f^{-1}((0, \infty))\).
      Since \((0, \infty)\) is open and \(f\) is a continuous function, then by continuous theorem preimage \(C = f^{-1}((0, \infty))\) is also open

      For any points \(p=(x_1,y_1)\) and \(q=(x_2,y_2)\), there are three segments can ensure that they are connected

      1. Segment \(pq\)
      2. If segment \(pq\) cannot connect them on \(A\), then consider two segment: 1. segment from \(p=(x_1,y_1)\) to \((x_{2},y_{1})\) 2. segment from \((x_{2},y_{1})\) to \(q=(x_2,y_2)\)
      3. If it doesn't work, then consider three segment: 1. segment from \(p=(x_{1},y_{1})\) to \((x_1,-1)\) 2. segment from \((x_1,-1)\) to \((x_2,-1)\) 3. segment from \((x_2,-1)\) to \(q=(x_{2},y_{2})\)

      Then we can also find a path

      image​

      Thus it's path connected. Thus \(C\) is a domain.

  2. (30 points)
    Compute (if they exist) the following limits:

    1. \(\lim_{(x,y) \to (0,0)} \frac{3x^2y}{x^4+y^2}\)
      Let \(y=x^2\), \(y=2x^2\)
      \(\lim_{(x,y) \to (0,0)} \frac{3x^2y}{x^4+y^2} = \lim_{(x,y) \to (0,0)} \frac{3x^4}{2x^4} = \frac{3}{2}\).
      \(\lim_{(x,y) \to (0,0)} \frac{3x^2y}{x^4+y^2} = \lim_{(x,y) \to (0,0)} \frac{6x^4}{5x^4} = \frac{6}{5}\).
      Thus limit doesn't exist.
    2. \(\lim_{(x,y) \to (0,0)} \frac{3x^3y^2}{x^2+y^2}\)
      Let \(x=r\cos\theta\), \(y=r\sin\theta\). Then \(= \lim_{r \to 0^+} \frac{3r^3\cos^3\theta r^2\sin^2\theta}{r^2} = \lim_{r \to 0^+} 3r^3 \cos^3\theta \sin^2\theta = 0\).
    3. \(\lim_{(x,y) \to (0,0)} \frac{y^3x}{x^2+y^6}\)
      Let \(x=y^3\). Then \(= \lim_{(x,y) \to (0,0)} \frac{y^6}{2y^6} = \frac{1}{2}\).
      \(x=2y^3\), then \(= \lim_{(x,y) \to (0,0)} \frac{2y^6}{5y^6} = \frac{2}{5}\). Thus limit doesn't exist.

  3. (10 points)
    Use the \(\epsilon-\delta\) definition to prove that \(\lim_{(x,y) \to (0,0)} \frac{x^5+y^5}{x^2+y^2} = 0\).
    We know \(|x| \le \sqrt{x^2+y^2}\), \(|y| \le \sqrt{x^2+y^2}\).
    Then \(|\frac{x^{5}+y^{5}}{x^{2}+y^{2}}| \le \frac{|x|^{5}+|y|^{5}}{x^{2}+y^{2}}= \frac{|x|^{3}x^{2}+|y|^{3}y^{2}}{x^{2}+y^{2}} \le \frac{(\sqrt{x^2+y^2})^{3} x^{2} + (\sqrt{x^2+y^2})^{3} y^{2}}{x^{2}+y^{2}}= \frac{(\sqrt{x^2+y^2})^{3} (x^{2}+y^{2})}{x^{2}+y^{2}}= (\sqrt{x^{2}+y^{2}})^{3}\).
    Since \(\sqrt{x^2+y^2} < \delta\), then this is \(< \delta^3\). We want \(\delta^3 \le \epsilon\), so take \(\delta = \epsilon^{1/3}\).

  4. (30 points)
    Prove or disprove that the following function is continuous at \((0,0)\)
    \(f(x,y) = \begin{cases} \frac{\sin(x^{2}+y^{2}) \cdot e^{-1/(x^2+y^2)}}{x^{2}+y^{2}} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0). \end{cases}\)
    NTP: \(\lim_{(x,y) \to (0,0)} \frac{\sin(x^2+y^2) \cdot e^{-\frac{1}{x^2+y^2}}}{x^2+y^2} = 0\).
    Since \(\lim_{(x,y) \to (0,0)} \frac{\sin(x^2+y^2) \cdot e^{-\frac{1}{x^2+y^2}}}{x^2+y^2} = \lim_{(x,y) \to (0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2} \cdot \lim_{(x,y) \to (0,0)} e^{-\frac{1}{x^2+y^2}}\).
    Then we prove \(\lim_{(x,y) \to (0,0)} e^{-\frac{1}{x^2+y^2}} = 0\). Let \(\sqrt{x^2+y^2} < \delta\), then \(\frac{1}{x^2+y^2} > \frac{1}{\delta^2}\).
    Thus \(\forall \varepsilon\), \(\exists\delta<\sqrt{\frac{1}{\ln(1/\epsilon)}}\) such that \(|e^{-\frac{1}{x^{2}+y^{2}}}| < |e^{-\frac{1}{\delta^{2}}}| < \epsilon\). Thus \(\lim_{(x,y) \to (0,0)} e^{-\frac{1}{x^2+y^2}} = 0\).
    Then \(\lim_{(x,y) \to (0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2} \cdot \lim_{(x,y) \to (0,0)} e^{-\frac{1}{x^2+y^2}} = \lim_{(x,y) \to (0,0)} \frac{\sin(x^2+y^2) \cdot e^{-\frac{1}{x^2+y^2}}}{x^2+y^2} = 0\).

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