For \(x>0\), let \[ f(x)=x\log(1+x^{-1}). \]
- Show that \(f\) is strictly increasing.
- Compute \[ \lim_{x\to0^+}f(x) \qquad\text{and}\qquad \lim_{x\to\infty}f(x). \]
Solution
1. For every \(u>0\),\[ \log(1+u)>\frac{u}{1+u}. \]
Define \[ h(u) \coloneqq \log(1+u)-\frac{u}{1+u}. \] Then \(h(0)=0\) and \[ h'(u) = \frac1{1+u}-\frac1{(1+u)^2} = \frac{u}{(1+u)^2} > 0 \] for \(u>0\). Hence \(h(u)>0\) for every \(u>0\).
2. The function \(f\) is strictly increasing on \((0,\infty)\).
Differentiating gives \[ \begin{aligned} f'(x) &= \log(1+x^{-1}) +x\frac{-x^{-2}}{1+x^{-1}}\\ &= \log(1+x^{-1})-\frac1{x+1}. \end{aligned} \] Set \(u=1/x>0\). Then \[ \frac1{x+1} = \frac{u}{1+u}, \] so step 1 gives \(f'(x)>0\).
3. One has\[ \boxed{ \lim_{x\to0^+}f(x)=0 }. \]
With \(u=1/x\), \[ f(x) = \frac{\log(1+u)}u, \] and \(x\to0^+\) is equivalent to \(u\to\infty\). By l’Hôpital’s rule, \[ \lim_{u\to\infty}\frac{\log(1+u)}u = \lim_{u\to\infty}\frac1{1+u} = 0. \]
4. One has\[ \boxed{ \lim_{x\to\infty}f(x)=1 }. \]
Again set \(u=1/x\). Then \(u\to0^+\) and \[ f(x) = \frac{\log(1+u)}u. \] Since \[ \lim_{u\to0}\frac{\log(1+u)}u=1, \] the desired limit is \(1\).
5. Q.E.D.
Step 2 proves Part (1), and steps <1>3–<1>4 give the two limits in Part (2).