\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

2017.2.7 Question 7

1. Since \(\ln \) is an increasing function, for \(0 < x < 1\), we have \(\ln x < 0\), and \begin {align*} f(x) > x & \iff \ln f(x) > \ln x \\ & \iff \ln x^x > \ln x \\ & \iff x \ln x > \ln x \\ & \iff x < 1, \end {align*}

   which is true since \(0 < x < 1\).

   Notice that \begin {align*} x < g(x) < f(x) & \iff \ln x < \ln x^{f(x)} < \ln x^x \\ & \iff \ln x < x^x \ln x < x \ln x \\ & \iff 1 > x^x > x. \end {align*}

   The right inequality is shown by the previous part. For the left inequality, we have \begin {align*} 1 > x^x & \iff \ln 1 > x \ln x \\ & \iff 0 > x \ln x \end {align*}

   must be true, since \(0 < x < 1\) and \(\ln x < 0\).

   Hence, we have \(x < g(x) < f(x)\) for \(0 < x < 1\).

   When \(x > 1\), we claim that \(x < f(x) < g(x)\).

2. Notice that \begin {align*} f'(x) & = \DiffOp {x} x^x \\ & = \DiffOp {x} \exp (x \ln x) \\ & = \exp (x \ln x) \cdot \left (1 \cdot \ln x + x \cdot \frac {1}{x}\right ) \\ & = \exp (x \ln x) \cdot (\ln x + 1) \\ & = f(x) \cdot (\ln x + 1). \end {align*}

   \(f'(x) = 0\) if and only if \(\ln x + 1 = 0\), which holds if and only if \(x = \frac {1}{e}\).

3. We have \[ \lim _{x \to 0} f(x) = \lim _{x \to 0} \exp (x \ln x) = \exp (0) = 1, \] and hence \[ \lim _{x \to 0} g(x) = 0. \]

4. Let \(h(x) = \frac {1}{x} + \ln x\). We have \[ h'(x) = - \frac {1}{x^2} + \frac {1}{x} = \frac {x - 1}{x^2}. \]

   When \(0 < x < 1\), \(h'(x) < 0\), and when \(1 < x\), \(h'(x) > 0\). Hence, \(h\) takes a minimum when \(x = 1\), and \(h(1) = \frac {1}{1} + \ln 1 = 1\).

   This shows precisely that \[ \frac {1}{x} + \ln x \geq 1 \] for \(x > 0\).

   Notice that \begin {align*} g'(x) & = \DiffOp {x} x^{f(x)} \\ & = \DiffOp {x} \exp (f(x) \ln x) \\ & = \exp (f(x) \ln x) \cdot \left (\frac {1}{x} \cdot f(x) + f'(x) \ln x\right ) \\ & = g(x) \cdot \left (\frac {1}{x} \cdot f(x) + f(x) \cdot (\ln x + 1) \cdot \ln x\right ) \\ & = f(x) g(x) \cdot \left (\frac {1}{x} + \ln x (\ln x + 1)\right ) \\ & \geq f(x) g(x) \cdot \left (1 + (\ln x)^2\right ) \\ & > 0, \end {align*}

   since \(f(x), g(x) > 0\) for \(x > 0\) (since they are both exponentials), and \(1 + (\ln x)^2 \geq 1 > 0\) as well.

The graphs of the functions look as follows.