# Bounds of $f(x)=(1-ax)^{1/x}$

Let $a\in(0,1)$ be a fixed number. What is the numeric value of upper and lower bound of $f(x)=(1-ax)^{1/x}$ on $x\in (0,1)$?

I feel as though I'm missing something, because it shouldn't be difficult. But trying to find $x$ for which $f'(x)=0$ is a bad idea to put it mildly and there has to be other way to see, whether $f$ has an extremum or not.

So far all I know is that

$$\lim_{x\to 0}(1-ax)^{1/x}=\lim_{x\to 0}e^{\frac{\log{(1-ax)}}{x}}=e^{\lim_{x\to 0}\frac{\log{(1-ax)}}{x}}\stackrel{H}{=}e^{\lim_{x\to 0}\frac{-a}{1-ax}}=e^{-a}$$

$$\lim_{x\to 1}(1-ax)^{1/x}=1-a$$

What do you propose to do now? (I hope the first limit is correct)

To find the derivative of $f$, use logarithmic differentiation: $$\ln(f(x)) = \ln[(1-ax)^{1/x}] = \frac{\ln(1-ax)}{x}\\ \frac{f'(x)}{f(x)} = \frac{-ax/(1-ax) - \ln(1-ax)}{x^2} = \frac{-ax - (1-ax)\ln(1 - ax)}{x^2(1-ax)}$$ If we can show that $f'$ never changes sign, we can guarantee that $f$ will be between its limits at $0$ and $1$. Noting that $f(x)$ is positive on $(0,1)$, it suffices to show that the numerator of the above is never zero, so that $f'/f$ never changes sign.
It may help to rewrite the numerator in the form $$(1-ax)[1-\ln(1-ax)]-1$$
• You seem to be missing an application of the chain rule in your differentiation of the logarithm. The numerator should be $-ax - (1-ax)\ln(1-ax)$. – Roger Burt Aug 12 '14 at 16:09
• Ok, but I still can't quite grasp why it doesn't change the sign. All I can say is that $1-\log{(1-ax)}>1$ (right?) – Jules Aug 13 '14 at 9:39
Bernoulli's Inequality, proven in this answer, says that for $ax\le1$ and $\frac yx\ge1$, $$1-\frac yxax\le(1-ax)^{y/x}$$ Therefore, for $ax,ay\le1$ and $0\lt x\le y\le1$, $$(1-ay)^{1/y}\le(1-ax)^{1/x}$$ This means that $f(x)=(1-ax)^{1/x}$ is a decreasing function of $x$. Therefore, for $x\in(0,1]$, $$1-a\le(1-ax)^{1/x}\le e^{-a}$$