Show that, for every sufficiently small \(\varepsilon>0\), the equation \[ x\left(1+\log\frac1{\varepsilon\sqrt{x}}\right)=1, \qquad x>0, \] has exactly two solutions. Let \(x(\varepsilon)\) be the smaller solution. Show that \[ x(\varepsilon)\to0 \qquad(\varepsilon\to0^+), \] but for every \(s>0\), \[ \varepsilon^{-s}x(\varepsilon)\to\infty \qquad(\varepsilon\to0^+). \]
Solution
For \[ 0<\varepsilon<1, \] set \[ L=\log\frac1\varepsilon>0 \] and define \[ F_\varepsilon(x) = x \left( 1+L-\frac12\log x \right), \qquad x>0. \] The equation in the problem is \[ F_\varepsilon(x)=1. \]
1. One has\[ \lim_{x\to0^+}F_\varepsilon(x)=0. \]
Write \[ F_\varepsilon(x) = (1+L)x-\frac12x\log x. \] The first term tends to zero, and the standard limit \[ x\log x\longrightarrow0 \qquad (x\to0^+) \] gives the same for the second term.
2. One has\[ \lim_{x\to\infty}F_\varepsilon(x)=-\infty. \]
Factor out \(x\): \[ F_\varepsilon(x) = x \left( 1+L-\frac12\log x \right). \] The factor in parentheses tends to \(-\infty\) as \(x\to\infty\), and is eventually negative. Multiplication by the positive quantity \(x\to\infty\) therefore sends the product to \(-\infty\).
3. The derivative is\[ F_\varepsilon'(x) = L+\frac12-\frac12\log x. \]
Differentiate \[ x \left( 1+L-\frac12\log x \right): \] \[ \begin{aligned} F_\varepsilon'(x) &= 1+L-\frac12\log x-\frac12\\ &= L+\frac12-\frac12\log x. \end{aligned} \]
4. The derivative vanishes at exactly one point,\[ \boxed{ x_* = e^{2L+1} = e\varepsilon^{-2}. } \]
By step 3, \[ F_\varepsilon'(x)=0 \] if and only if \[ \log x=2L+1. \] Exponentiating gives \[ x=e^{2L+1}. \] Since \[ e^L=\varepsilon^{-1}, \] this equals \(e\varepsilon^{-2}\).
5. The function \(F_\varepsilon\) is strictly increasing on\[ (0,x_*) \]and strictly decreasing on\[ (x_*,\infty). \]
The expression in step 3 is positive exactly when \[ \log x<2L+1, \] equivalently \(x<x_*\), and negative exactly when \(x>x_*\).
6. The maximum value of \(F_\varepsilon\) is\[ \boxed{ F_\varepsilon(x_*) = \frac{e}{2\varepsilon^2}. } \]
By step 4, \[ \log x_*=2L+1. \] Hence \[ \begin{aligned} F_\varepsilon(x_*) &= x_* \left( 1+L-\frac12(2L+1) \right)\\ &= \frac{x_*}{2}\\ &= \frac{e}{2\varepsilon^2}. \end{aligned} \]
7. For every sufficiently small \(\varepsilon>0\), the equation\[ F_\varepsilon(x)=1 \]has exactly two positive solutions.
Choose \(\varepsilon\) so small that \[ \frac{e}{2\varepsilon^2}>1. \] By steps 1, 5, and 6, the function increases strictly from the limiting value \(0\) at the left endpoint to a value greater than \(1\) at \(x_*\). The intermediate value theorem and strict monotonicity therefore give exactly one solution in \((0,x_*)\).
By steps 2, 5, and 6, the function then decreases strictly from a value greater than \(1\) at \(x_*\) to \(-\infty\). Hence there is exactly one solution in \((x_*,\infty)\). These are the only positive solutions.
8. Let \(x(\varepsilon)\) denote the smaller solution. For every fixed\(\delta>0\),\[ F_\varepsilon(\delta)\longrightarrow\infty \qquad (\varepsilon\to0^+). \]
For fixed \(\delta>0\), \[ F_\varepsilon(\delta) = \delta \left( 1+\log\frac1\varepsilon -\frac12\log\delta \right). \] The only term depending unboundedly on \(\varepsilon\) is \[ \delta\log\frac1\varepsilon, \] which tends to \(+\infty\).
9. The smaller solution satisfies\[ \boxed{ x(\varepsilon)\longrightarrow0 } \]as \(\varepsilon\to0^+\).
Fix \(\delta>0\). Since \[ x_*=e\varepsilon^{-2}\longrightarrow\infty, \] one has \(\delta<x_*\) for all sufficiently small \(\varepsilon\). By step 8, also \[ F_\varepsilon(\delta)>1 \] for all sufficiently small \(\varepsilon\).
On \((0,x_*)\), the function is strictly increasing by step 5, and the smaller root is the unique point there at which \[ F_\varepsilon(x)=1. \] Thus \[ 0<x(\varepsilon)<\delta \] for all sufficiently small \(\varepsilon\). Since \(\delta>0\) was arbitrary, the claimed limit follows.
10. Fix\[ s>0 \qquad\text{and}\qquad M>0. \]Then\[ F_\varepsilon(M\varepsilon^s) \longrightarrow0 \qquad (\varepsilon\to0^+). \]
Since \[ \log(M\varepsilon^s) = \log M-sL, \] one has \[ \begin{aligned} F_\varepsilon(M\varepsilon^s) &= M\varepsilon^s \left( 1+L-\frac12\log M+\frac{s}{2}L \right)\\ &= M\varepsilon^s \left( 1-\frac12\log M + \left( 1+\frac{s}{2} \right)L \right). \end{aligned} \] The standard limit \[ \varepsilon^s\log\frac1\varepsilon\longrightarrow0 \] shows that the right-hand side tends to zero.
11. For every fixed \(s>0\) and \(M>0\), one has\[ x(\varepsilon)>M\varepsilon^s \]for all sufficiently small \(\varepsilon>0\).
5. Since\[ F_\varepsilon(M\varepsilon^s)<1 = F_\varepsilon(x(\varepsilon)), \]strict monotonicity gives\[ M\varepsilon^s<x(\varepsilon). \]
12. For every \(s>0\),\[ \boxed{ \varepsilon^{-s}x(\varepsilon) \longrightarrow \infty } \]as \(\varepsilon\to0^+\).
Fix \(s>0\). Step 11 says that for every \(M>0\), \[ \frac{x(\varepsilon)}{\varepsilon^s}>M \] for all sufficiently small \(\varepsilon\). This is exactly divergence of the ratio to \(+\infty\).
13. Q.E.D.
Step 7 proves the two-solution assertion, step 9 proves the first limit, and step 12 proves the second limit.