I'm trying to do the following problem and could use some help(from Apostol, Calculus, Volume I, 7.11 Ex. 33 p. 291):

A funtion $f$ has a continuous third derivative everywhere and satisfies the relation

$$ \lim_{x \to 0} \left(1 + x + \dfrac{f(x)}{x}\right)^{1/x} = e^3.$$

Compute $f(0), f'(0), f''(0),$ and $\lim_{x \to 0} \left(1 + \frac{f(x)}{x}\right)^{1/x}$.

[Hint: If $\lim_{x \to 0} g(x) = A$, then $g(x) = A + o(1)$ as $x \to 0$.]

The book gives the following as answers: $f(0) = 0, f'(0) = 0, f''(0) = 4$ and the limit $= e^2$.

I cannot seem to make any forward progress on this. It is the last Exercise in a section of exercises on taking limits by using polynomial expansions of functions. (This set of Exercises is immediately before the section on L'Hopital's rule... I don't know if that is applicable here, but a solution without it is appreciated since Apostol intends this to be done without it.)

My initial attempts involve writing:

$$\begin{align*} & \lim_{x \to 0} \left(1 + x + \dfrac{f(x)}{x}\right)^{1/x} &= e^3.\\ \implies & \lim_{x \to 0} \left(e^{(1/x)\log(1 + x + f(x)/x)}\right) &= e^3.\\ \implies & \lim_{x \to 0} \left(e^{-\frac{\log x}{x} + \frac{1}{x} \log (x + x^2 + f(x))}\right) &= e^3. \end{align*}$$

I wanted to do this to attempt to get to a point that I could write a polynomial expansion of $\log$ at $0$, but I can't seem to make any progress on that front. Maybe there is a better way to simplify things?

Thanks for any help. Full solutions or hints are equally welcome.

  • $\begingroup$ @PeterT.off It was some error in your typing. $\endgroup$ – AD. Apr 10 '12 at 20:56
  • 1
    $\begingroup$ What is going on with this question. I can't see any formulas only here, but not in other questions! $\endgroup$ – Norbert Apr 10 '12 at 21:04
  • $\begingroup$ @PeterT.off: I might well be that because of your comment, all the math on this site is broken. Could you please try to delete your comment? $\endgroup$ – Fabian Apr 10 '12 at 21:25
  • $\begingroup$ @Fabian Really? How is that even possible? $\endgroup$ – Pedro Tamaroff Apr 10 '12 at 21:28
  • $\begingroup$ @PeterT.off: don't know how this is possible. I just know that now the site is fine again (if you remember the content of your comment then maybe you should start a thread on meta) $\endgroup$ – Fabian Apr 10 '12 at 21:29

The hypothesis is that $\left(1+x+x^{-1}f(x)\right)^{1/x}=\mathrm e^{3+o(1)}$, that is, $1+x+x^{-1}f(x)=\mathrm e^{3x+o(x)}$. Since $\mathrm e^{3x+o(x)}=1+3x+o(x)$, one gets $x^{-1}f(x)=2x+o(x)$. Thus $f(x)=2x^2+o(x^2)$.

Assume that $f$ is twice differentiable at $0$. Then $f(x)=f(0)+f'(0)x+\frac12f''(0)x^2+o(x^2)$, by Taylor's theorem. By the uniqueness of the coefficients of the expansion of $f$ at $0$, one gets $f(0)=f'(0)=0$ and $f''(0)=4$.

Note that the fact that $f$ is twice differentiable at $0$ is a hypothesis, and certainly not a consequence of the expansion of $f$ at $0$. (On the other hand, the hypothesis that $f$ has a continuous third derivative everywhere is stronger than needed.)

Finally, $x^{-1}f(x)=o(1)$ hence $\log(1+x^{-1}f(x))=x^{-1}f(x)+o(x^{-1}f(x))=2x+o(x)$ and $x^{-1}\log(1+x^{-1}f(x))=2+o(1)$.

Taking exponentials of both sides yields $\left(1+x^{-1}f(x)\right)^{1/x}=\mathrm e^{2+o(1)}$, which is equivalent to the statement that $\left(1+x^{-1}f(x)\right)^{1/x}\to\mathrm e^{2}$ when $x\to0$.

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