12.6 Bolzano, Intermedia value theorem

tutorial 8 part 2.pdf

1. Prove if the following functions are injective and surjective, then if they are bijective
   compute its inverse.

   1. \(f(x) = (x - 1)^2\)

      \(f: \mathbb{R} \to \mathbb{R},\) continuous

      \(\text{Im} f = [0, +\infty)\) thus \(f\) is not injective in \(\mathbb{R}\) and not surjective in \(\mathbb{R}\).

      Define \(g: [1, +\infty) \to [0, +\infty)\)

      \(g\) is injective and surjective, continuous. \(\therefore \exists \, g^{-1}: [0, +\infty) \to [1, +\infty)\) and \(g\) is continuous.

      \(f(x) = (x - 1)^2 \implies 0 \leq y = (x - 1)^2\)

      \(\sqrt{y} = |x - 1| \implies x - 1 = \sqrt{y} \implies x = \sqrt{y} + 1\)

      Because \(x \geq 1\), \(g^{-1}(y) = \sqrt{y} + 1\)

      Verification: \(g \circ g^{-1}(y) = y\) and \(g^{-1} \circ g(x) = x\)

      \(\therefore g^{-1}(x) = \sqrt{x} + 1\) 2. \(g(x) = \ln(x) + 1\)​

      \(g: (0, +\infty) \to \mathbb{R},\) continuous in \((0, +\infty)\)

      Suppose \(g(x_1) = g(x_2)\), then \(\ln(x_1) + 1 = \ln(x_2) + 1\), then \(\ln(x_1) = \ln(x_2)\), thus \(x_1=x_2\)

      Thus, \(g\) is injective.

      For any \(y \in \mathbb{R}\), we must find \(x \in (0, +\infty)\) such that \(g(x) = y\)

      Since \(\ln(x) + 1 = y\), then \(\ln(x) = y - 1\) which is \(x = e^{y - 1}\)

      \(\therefore g\) is surjective.

      Since \(g\) is injective and surjective, it is bijective. \(\therefore \exists \, g^{-1}: \mathbb{R} \to (0, +\infty)\) and \(g\) is continuous.

      \(g^{-1}(x)=e^{x-1}\) 3. \(t(x) = e^{2x} - 5\)

      \(t: \mathbb{R} \to \mathbb{R},\) continuous and \(e^x\) is continuous

      Then \(e^{2x} = \exp(2x)\) is the composition of continuous functions, thus \(t(x)\) is continuous.

      Suppose \(t(x_1) = t(x_2)\), then \(e^{2x_1} - 5 = e^{2x_2} - 5\), then \(e^{2x_1} = e^{2x_2}\), then \(\ln(e^{2x_1}) = \ln(e^{2x_2})\)

      Thus \(x_1=x_2\)

      Thus, \(t\) is injective.

      For any \(y \in \mathbb{R}\), we must find \(x \in \mathbb{R}\) such that \(t(x) = y\)

      Since \(e^{2x} - 5 = y\), then \(x = \frac{1}{2} \ln(y + 5)\).

      Thus, \(t\) is surjective.

      Since \(t\) is injective and surjective, it is bijective. \(\therefore \exists \, t^{-1}: \mathbb{R} \to \mathbb{R}\) and \(t\) is continuous.

      \(t^{-1}(x)=\frac12\ln(x+5)\)

2. Compute the following limits:

   1. \(\lim_{x \to \infty} \frac{\ln(x)}{x}\)

      Let \(\lim_{x\to\infty}\frac{\ln(x)}{x}=\lim_{e^{x}\to\infty}\frac{\ln(e^{x})}{e^{x}}=\lim_{x\to\infty}\frac{x}{e^{x}}\).

      Since \(0<\frac{x}{e^{x}}<\frac{2^{x}}{e^{x}}\) and \(\lim\frac{2^{x}}{e^{x}}=0\), then \(\lim_{x \to \infty} \frac{\ln(x)}{x}=0\)​ 2. \(\lim_{x \to 0^+} x^x\)

      Since \(x^{x}=e^{\ln x^{x}}=e^{x\ln x}=e^{\frac{\ln x}{\frac{1}{x}}}=e^{-\frac{\ln\frac{1}{x}}{\frac{1}{x}}}\), then \(\lim_{x\to0^{+}}x^{x}=\lim_{x\to0^{+}}e^{-\frac{\ln\frac{1}{x}}{\frac{1}{x}}}=e^{\lim_{x\to0^{+}}-\frac{\ln\frac{1}{x}}{\frac{1}{x}}}=e^0=1\)​