I have a question about evaluating the limit:

$$\lim_{x \to\infty }\left(x^{f(x)}-x \right)$$


$f(x)$ is a continuous map from the positive reals to the positive reals , and

$\lim_{x\rightarrow \infty }f(x)= 1$.

I attempted to apply L'Hôpital's rule by writing:

$x^{f(x)}-x$ = $\log(\exp(x^{f(x)})/\exp(x))$

then applying the rule to $\exp(x^{f(x)})/\exp(x)$, however this quotient appeared in the resulting expression and successive applications of the rule would not remove it.

The Wikipedia article on L'Hôpital's rule (link below) mentions the way the original expression can occur in the result of applying the rule. The article gives some examples where this problem is solved by using transformations but I could not get that method to work in this case.

I would appreciate any help in evaluating this limit and/or in referring me to a source where it or similar limits are evaluated.



Thanks for the comment and the answer (now deleted). They show me I left some information out of my question. Apologies for the omission. I should have included the following:

  1. The function $f(x)$ is assumed to be $C^\infty$ on the positive reals.

  2. The limit will depend on $f(x)$ so I was looking for an evaluation of the limit that related the limit to the properties of $f(x)$. For example, I was looking for those properties of $f(x)$ that imply the limit is $\infty$ and those that imply it is finite.

  • 1
    $\begingroup$ $f$ is not necessairily diferentiable, when applying the chain rule using L'Hopital, you are implying differentiability. $\endgroup$ – Joaquin Liniado May 26 '15 at 23:28
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    $\begingroup$ As is easily shown using the approach in the deleted answer, if $(f(x)-1)\ln x\to c$ then $x^{f(x)}-x\to e^c-1$. Of course, if $(f(x)-1)\ln x\to+\infty$ then $x^{f(x)}-x\to+\infty$. And if $(f(x)-1)\ln x$ diverges because it oscillates then so does $x^{f(x)}-x$. $\endgroup$ – Did May 27 '15 at 20:42
  • $\begingroup$ Thanks for the comment but it seems have an incorrect implication. If $f(x)$ has the property $(f(x)-1)\ln x\to c$ then $x^{f(x)}-x\to x(e^c-1)$ and not $x^{f(x)}-x\to e^c-1$. $\endgroup$ – gjh May 27 '15 at 21:22
  • $\begingroup$ Your approach is to start with $(f(x)-1)\ln x\to c$ to derive the behavior of $\lim_{x \to\infty }\left (x^{f(x)}-x \right )$ but $(f(x)-1)\ln x\to c$ is equivalent to $x^{f(x)-1}\to e^c$ so this approach only shifts the question to what properties of $F(x)$ determine the behavior of $\lim_{x \to\infty }\left (x^{F(x)}\right )$ where $F(x) = f(x) – 1$ and $\lim_{x\rightarrow \infty }F(x)= 0$. $\endgroup$ – gjh May 27 '15 at 22:26
  • $\begingroup$ *Read $(f(x)-1)x\ln x\to c$ and $x^{f(x)}-x\to c$. (Unrelated: Please use @.) $\endgroup$ – Did May 28 '15 at 9:22

Let $L$ the desired limit. Take $f(x)=1$ we get trivially $L=0$. Now take $f(x)=1+\frac1{\ln x}$ we get

$$x^{f(x)}-x=x\left(x^{f(x)-1}-1\right)=x\left(e-1\right)\xrightarrow{x\to\infty}\infty$$ so we see that the result depends on the choice of $f$.


I now think the problem is simpler than it originally seemed to me when I posted the question.

Is the following how the form of $f(x)$ determines $\lim_{x \to\infty }\left(x^{f(x)}-x \right)$?

CASE ONE : $\lim_{x \to\infty }\left(x^{f(x)}-x \right) = L$ (finite)

This limit implies $x^{f(x)}-x = L + g(x)$

where $\lim_{x\rightarrow \infty }g(x)= 0$ and $L$ is a constant.

Rearranging $x^{f(x)}-x = L + g(x)$ gives:

$f(x) = 1+[log[1+[L+g(x)]/x]/log(x)$

The implications are reversible so:

$\lim_{x \to\infty }\left(x^{f(x)}-x \right) = L$ (finite) iff $f(x) = 1+[log[1+[L+g(x)]/x]/log(x)$

provided $\lim_{x\rightarrow \infty }g(x)= 0$

CASE TWO : $\lim_{x \to\infty }\left(x^{f(x)}-x \right) = \infty$

This limit implies:

$x^{f(x)}-x = h(x)$

where $\lim_{x\rightarrow \infty }h(x)= \infty$

Rearranging $x^{f(x)}-x = h(x)$ gives:

$f(x) = 1+[log[1+h(x)/x]/log(x)$

The implications are reversible so:

$\lim_{x \to\infty }\left(x^{f(x)}-x \right) = \infty$ iff $f(x) = 1+[log[1+h(x)/x]/log(x)$

provided $\lim_{x\rightarrow \infty }h(x)= \infty$



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