Problem: $f: \mathbb{R^{+}} \to \mathbb{R^{+}}$ and strictly positive. Given that:

$$\lim_{h \to 0} \left(\frac{f(x_0+h)}{f(x_0)}\right)^{\frac{1}{h}}= L,$$

where $L$ is not $0$ and is finite. Show that $f$ is differentiable at $x_0$. Nothing else is given about $f$, not even continuity.

My attempt: The only approach that I can think of is trying to show that $\lim_{h \to 0} \left(\frac{f(x_0+h)-f(x_0)}{h}\right)$ exists. Now, if I assume that $f$ is continous at $x_0$ then the problem becomes easy, because the given condition would then give $f(x_0)=\lim_{h \to 0}f(x_0+h)=x_0$ and $\lim_{h \to 0}e^{\frac{f(x_0+h)-x_0}{hx_0}}=L$. Which means $\frac{f(x_0+h)-f(x_0)}{hx_0}$ exists and so does $\frac{f(x_0+h)-f(x_0)}{hx_0}$. But continuity is not given :(.

I think (not sure) we need to define some function which is related to the things above and is continous (continuous extension of some function above) but I don't know how to do that in this case.

Someone please help.

  • $\begingroup$ can't you write down a counterexample if $f$ is not continuous? $\endgroup$ – Julian Braun Apr 15 '16 at 8:07
  • $\begingroup$ @JulianBruan, oops! yes. $f(x)=x$ for x not equal to 5 and $f(5)=6$ $\endgroup$ – Subham Jaiswal Apr 15 '16 at 8:21
  • $\begingroup$ @JulianBraun, now should I delete this question? what should I do now? $\endgroup$ – Subham Jaiswal Apr 15 '16 at 8:22

@Cupid, Don't delete the question; rather, it is almost certain you have a typo: it should be $f(x_0+h)/f(x_0)$, with $f(x_0)$ in the denominator, not $x_0$.

And then you were / are on the right track. $e^t$ and $\ln t$ are continuous everywhere. If the limit exists and is finite and strictly greater than 0, this means $$ \lim_{h \to 0} \frac{\ln f(x_0+h) - \ln f(x_0)} h = \ln L$$ This limit exists and is finite, so $\ln f(x)$ is differentiable at $x_0$. Since the exponential function is differentiable everywhere, this means $f(x) = e^{\ln f(x)}$ is differentiable at $x_0$.

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