Skip to content

11.29 Continuity of functions

\(f\) is continuous in \(x = a\) if \(\forall \epsilon>0\), \(\exists \delta > 0\) : \(\forall x \in\) Dom \(f,\) \(|x-a| \leq \delta\) it follows that \(|f(x) - f(a)| < \epsilon\)

Limit point: \((a-\delta,a) \cup (a,a+\delta)\) Continuous: \((a-\delta, a+\delta)\)

\(f\) is not continuous at \(x_0\) if \(\lim_{x \to x_0^-}f(x) = \lim_{x \to x_0^+}f(x) = L \Rightarrow \lim_{x \to x_0}f(x) = L \neq f(x_0)\)

image image​

  1. Prove \(f(x) = \begin{cases} \frac{x+1}{x-1} & x < 0 \\ x\sin x & x \geq 0 \end{cases}\) is continuous or not image​

    Since \(y = \frac{x+1}{x-1}\) is continuous in \(x < 0\), then \(f(x)\) is continuous in \(x < 0\).

    \(y = x\sin x\) is continuous in \(x > 0\), since \(x\) and \(\sin x\) is continuous in \(x>0\), then \(f(x)\) is continuous at \(x > 0\)

    But \(\lim_{x \to 0^-}f(x) = -1 \neq 0 = \lim_{x \to 0^+}f(x) \Rightarrow \nexists \lim_{x \to 0}f(x)\) \(\Rightarrow f\) is discontinuous at \(x = 0\)

  2. \(f(x)=\lfloor x\rfloor=\text{ greatest integer }\leq x\)

    image​

    \(f(x)\) is continuous in \(\bigcup_{x \in \mathbb{Z}}(x,x+1)\) Dom \(f = \mathbb{R}\)
    ​\(\forall x_0 \in \mathbb{Z}\), \(\lim_{x \to x_0^-}f(x) = x_0 \neq x_0-1 = \lim_{x \to x_0^+}f(x)\) \(\Rightarrow \nexists \lim_{x \to x_0}f(x)\)
    ​\(\Rightarrow f(x)\) is discontinuous at every \(x \in \mathbb{Z}\)

  3. \(f\) is continuous in \((a,b)\)
    ​\(g\) is continuous in \((b,c)\)
    ​\(\lim_{x \to b^-}f(x) = \lim_{x \to b^+}g(x) = L\)

    image​

    Let \(h(x) = \begin{cases} f(x) & a < x < b \\ L & x = b \\ g(x) & b < x < c \end{cases}\)

    Prove \(h(x)\) is continuous on \((a,c)\)

    Since \(h(x)=f(x)\) in \((a,b)\) and \(f(x)\) is continuous in \((a,b)\), then \(h(x)\) is continuous in \((a,b)\)
    Since \(h(x)=g(x)\) in \((b,c)\) and \(g(x)\) is continuous in \((b,c)\), then \(h(x)\) is continuous in \((b,c)\)
    Need to check \(x = b\) is continuous, use definition

    Thus we need to prove \(|x-b| < \delta\), \(|h(x) - h(b)| < \epsilon\)
    Since \(\lim_{x \to b^-}f(x) = L\), then \(\exists \delta_1 > 0\) : \(\forall x \in\) Dom \(f\), \(x \in (b-\delta_1,b)\), \(|f(x)-L| < \epsilon\)
    Since \(\lim_{x \to b^+}g(x) = L\), then \(\exists \delta_2 > 0\) : \(\forall x \in\) Dom \(g\), \(x \in (b,b+\delta_2)\), \(|g(x)-L| < \epsilon\)
    Pick \(\delta = \min\{\delta_1,\delta_2\}\), \(|h(x)-L| < \epsilon\) for \(x \in (b-\delta,b) \cup (b,b+\delta)\)
    for \(x = b\) we have \(|h(b) - L| = 0 < \epsilon\)

    Then \(|h(x) - h(b)| < \epsilon\) for \(|x-b| < \delta\) \(\Rightarrow \lim_{x \to b}h(x) = h(b) = L\)

  4. \(f(x)=x\sin(\frac{1}{x})\) \(x\neq 0\), prove \(\lim_{x\to0}f\left(x\right)=0\)

    image​

    Since \(\lim_{x\to\infty}x\sin(\frac{1}{x})=\lim_{x\to\infty}\frac{\sin(\frac{1}{x})}{\frac{1}{x}}=1\), then \(y=1\) is an horizontal asymptote

    By exercise 3, we can construct a function \(h(x)=\begin{cases}f\left(x\right),x\neq0\\0,x=0\end{cases}\) which is continuous

    Since \(|\sin\theta|\leq1\Rightarrow-1\leq\sin\frac{1}{x}\leq1\Rightarrow-x\leq x\sin\left(\frac{1}{x}\right)\leq x\), consider two cases

    1. When \(x>0\), \(\lim_{x\to0^{+}}-x=0=\lim_{x\to0^{+}}x\) Then \(\lim_{x\to0^{+}}x\sin(1/x)=0\)​
    2. When \(x<0\), \(\lim_{x\to0^{-}}-x=0=\lim_{x\to0^{-}}x\) Then \(\lim_{x\to0^{-}}x\sin(1/x)=0\)

    Thus \(\lim_{x\to0}x\sin(1/x)=0\)

‍