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5.7 Metric Space

  1. Find the interior and boundary of the following sets. Are they open? bounded? connected?

    \(A=\{x\in \R^n:|x|\leq 1\}\)​

    Since \(A=\overline{B(0,1)}\) which is closed not open

    Since \(A\subseteq B(0,2)\), then it is bounded

    \(A\) is connected

    \(B=\{x\in \R^{n}:x_{i}\in \mathbb{Q}\}\)​

    \(B^{\circ}=\empty\) and \(\partial B=\R^n\) and \(B\) is not open and not bounded and not connected

Limits of two variable \(f:D\subseteq \R^2\to \R\)

If \(\forall \varepsilon>0,\exists \delta_\varepsilon>0\) such that \(0<||(x,y)-(x_0,y_0)||<\delta\), then \(|f(x,y)-L|<\varepsilon\), then \(\lim\limits_{(x,y)\to (x_0,y_0)}f(x,y)=L\)

  1. \(\lim\limits_{(x,y)\to (1,1)}\left[\frac{x^{2}-1}{x-1}+\frac{y-1}{y^{2}-1}\right] =\lim\limits_{(x,y)\to (1,1)}\frac{x^{2}-1}{x-1}+\lim\limits_{(x,y)\to (1,1)}\frac{y-1}{y^{2}-1} =2+\frac{1}{2}=\frac{5}{2}\)

  2. \(\lim\limits_{(x,y)\to(0,0)}\frac{\sin(x)\sin(3y)}{2xy}=\frac{1}{2}\lim\limits_{(x,y)\to(0,0)} \frac{\sin x}{x}\cdot3\frac{\sin(3y)}{3y}=\frac{3}{2}\)

  3. Prove that \(\lim\limits_{(x,y)\to (0,0)}\frac{x^3+y^3}{x^2+y^2}=0\), we need to show that \(\forall \varepsilon>0,\exists \delta>0\) such that if \(\sqrt{x^{2}+y^{2}}<\delta\), then \(\left|\frac{x^3+y^3}{x^2+y^2}-0\right|<\varepsilon\)

    First we bound \(|x^3+y^3|\)​

    Since \(|x^{3}|=|x|x^{2}=\sqrt{x^{2}}x^{2}\leq \sqrt{x^2+y^2}x^2\) and \(|y^3|\leq \sqrt{x^2+y^2}y^2\)​

    Now by triangular inequality, we get \(|x^{3}+y^{3}|\leq\sqrt{x^{2}+y^{2}}(x^{2}+y^{2})=(x^2+y^2)^\frac{3}{2}\)​

    Then \(\left|\frac{x^3+y^3}{x^2+y^2}\right|\leq\left|\frac{(x^2+y^2)^\frac{3}{2}}{x^2+y^2}\right|\leq\left|\sqrt{x^2+y^2}\right|<\delta =\varepsilon\)

  4. \(\lim\limits_{(x,y)\to (0,0)}\frac{x^{2}e^{-y^2}}{x^{4}+y^{2}}\)

    If the limit exists, then it must exist for all paths that goes through \((0,0)\)

    Along \(x=0\), then \(f(0,y)=\frac{0e^{-y^2}}{0+y^{2}}=0\)​

    Along \(y=x^2\), then \(f(x,x^{2})=\frac{x^{2}e^{-x^4}}{2x^{4}}=\frac{e^{-x^4}}{2x^{2}}\), then \(\lim_{x\to0}\frac{e^{-x^4}}{2x^{2}}=\lim_{x\to0}\frac{1}{2x^{2}\cdot e^{x^4}}=\infty\)​

    Thus the limit doesn't exist

  5. \(\lim\limits_{(x,y)\to (0,0)}\frac{x^{4}y}{x^{4}+y^{4}}\)​

    Let \(y=kx\), then \(f(x,kx)=\frac{kx^{5}}{x^{4}\left(1+k^{4}\right)}\) and the limit is \(0\) when \(x\to 0\)​

    If \(y=x^2\), then \(f(x,x^2)\) and the limit is 0 when \(x\to 0\)​

    Thus we use definition to prove this

    Since \(\left|\frac{x^{4}y}{x^{4}+y^{4}}\right|\leq \left|\frac{x^{4}y}{x^{4}}\right|=|y |=\sqrt{y^{2}}\leq \sqrt{x^{2}+y^{2}}\), then let \(\varepsilon >0\), take \(\delta=\varepsilon\), if \(||(x,y)||=\sqrt{x^2+y^2}<\delta\)

    Then \(\left|\frac{x^{4}y}{x^{4}+y^{4}}\right|<\delta=\varepsilon\)

  6. Define \(f(0,0)\) in a way that extends \(f(x,y)=\frac{xy(x^{2}-y^{2})}{x^{2}+y^{2}}\) to be continuous at \((0,0)\)​

    Since \(f\) is not defined at \((0,0)\) so we need to prove it's continuous at \((0,0)\)​

    Recall the squeeze theorem

    We need to find the limit of \(f\) at \((0,0)\)​

    In our case, \(|x|\leq \sqrt{x^2+y^2}\) and \(|y|\leq \sqrt{x^2+y^2}\) and \(|x^2-y^2|\leq |x^2+y^2|\)​

    Then \(\left|\frac{xy(x^{2}-y^{2})}{x^{2}+y^{2}}\right|\leq|x^2+y^2|\)

    So \(-(x^2+y^2)\leq f(x,y)\leq x^2+y^2\), then since \(\lim\limits_{(x,y)\to (0,0)}-x^2-y^2=0=\lim\limits_{(x,y)\to (0,0)}x^2+y^2\), then by squeeze theorem, we get \(\lim\limits_{(x,y)\to (0,0)}\frac{xy(x^{2}-y^{2})}{x^{2}+y^{2}}=0\)

Theorem: Let \(f: D \subseteq \mathbb{R}^2 \to \mathbb{R}\) where \(D\) is a neighborhood of \((0,0)\). Then \(\lim_{(x,y)\to(0,0)} f(x,y) = L\) if and only if the following two conditions hold:

  1. For all \(\theta \in [0, 2\pi)\), \(\lim_{r \to 0^+} f(r \cos \theta, r \sin \theta) = L\).

  2. The limit is uniform with respect to \(\theta\). This is \(\forall \epsilon > 0\), there exists \(\delta > 0\) such that \(|f(r \cos \theta, r \sin \theta) - L| < \epsilon\) for all \(r \in (0, \delta)\) and all \(\theta \in [0, 2\pi)\).

Example

  1. \(\lim_{(x,y)\to(0,0)} \frac{x^4y}{x^4+y^4}\)
    Let \(x = r \cos \theta, y = r \sin \theta\), then \(\lim_{r \to 0^+} \frac{(r \cos \theta)^4 (r \sin \theta)}{(r \cos \theta)^4 + (r \sin \theta)^4} = \lim_{r \to 0^+} \frac{r^5 \cos^4 \theta \sin \theta}{r^4 (\cos^4 \theta + \sin^4 \theta)} = \lim_{r \to 0^+} r \left( \frac{\cos^4 \theta \sin \theta}{\cos^4 \theta + \sin^4 \theta} \right) = \lim_{r \to 0^+} (\text{bounded})r = 0\)
    So \(\lim_{(x,y)\to(0,0)} \frac{x^4y}{x^4+y^4} = 0\)

  2. \(\lim_{(x,y)\to(0,0)} \frac{x^3y}{x^2+y^2}\)​

    Again \(\lim_{r \to 0^+} \frac{(r \cos \theta)^3 (r \sin \theta)}{(r \cos \theta)^2 + (r \sin \theta)^2} = \lim_{r \to 0^+} \frac{r^4 \cos^3 \theta \sin \theta}{r^2 (\cos^2 \theta + \sin^2 \theta)} = \lim_{r \to 0^+} r^2 (\cos^3 \theta \sin \theta) = \lim_{r \to 0^+} (\text{bounded}) r^2 = 0\)
    It doesn't depend on \(\theta\). Thus \(\lim_{(x,y)\to(0,0)} \frac{x^3y}{x^2+y^2} = 0\)