Let \(f:(0,\infty)\to(0,\infty)\) be differentiable. Prove that for every \(x>0\) the finite nonzero limit \[ \lim_{\delta\to0} \left(\frac{f(x+\delta x)}{f(x)}\right)^{1/\delta} \] exists.
Solution
Fix \(x>0\).
1. The function\[ h(t)\coloneqq\log f(t) \]is differentiable at \(x\), with\[ h'(x)=\frac{f'(x)}{f(x)}. \]
The hypothesis gives \(f(x)>0\), and \(f\) is differentiable at \(x\). Therefore the chain rule applies to \(\log\circ f\) and yields the displayed derivative.
2. One has\[ \lim_{\delta\to0} \frac{ \log f(x+\delta x)-\log f(x) }{\delta} = \frac{x f'(x)}{f(x)}. \]
For \(\delta\neq0\) sufficiently close to \(0\), the point \(x+\delta x=x(1+\delta)\) remains positive. Rewrite the quotient as \[ x\, \frac{ h(x+\delta x)-h(x) }{ \delta x }. \] As \(\delta\to0\), the second factor tends to \(h'(x)\) by the definition of the derivative. Apply step 1.
3. For every sufficiently small nonzero \(\delta\),\[ \left( \frac{f(x+\delta x)}{f(x)} \right)^{1/\delta} = \exp\left( \frac{ \log f(x+\delta x)-\log f(x) }{\delta} \right). \]
Both numerator and denominator inside the ratio are positive, so the real logarithm is defined. For every \(u>0\) and real \(r\), \[ u^r=e^{r\log u}. \] Apply this with \[ u=\frac{f(x+\delta x)}{f(x)} \qquad\text{and}\qquad r=\frac1\delta. \]
4. Therefore\[ \boxed{ \lim_{\delta\to0} \left( \frac{f(x+\delta x)}{f(x)} \right)^{1/\delta} = \exp\left(\frac{x f'(x)}{f(x)}\right) }. \]
2. Since the exponential function is continuous, step 2 givesthe displayed limit.
5. The limit in step 4 is finite and nonzero.
The number \[ \frac{x f'(x)}{f(x)} \] is finite because \(f\) is differentiable and \(f(x)>0\). Its exponential is therefore a finite positive real number.
6. Q.E.D.
Steps 4 and 5 prove the required existence, finiteness, and nonvanishing.