# What's wrong in this evaluation $\lim_{x\to\infty}x^{\frac{1}{x}}$ and why combinatorial arguments cannot be made?

The answers (as well as the premise) of this question have me confused. Please point out the error: $$\displaystyle \lim_{x\to\infty}x^{\frac{1}{x}}=\lim_{x\to\infty}e^{\frac{\log x}{x}}$$

By L' Hospital rule $$\exp\left( \lim_{x\to\infty} \frac{\log x}{x}\right) = e^0 =1$$

Three of the answers state an example of the form $x\cdot\frac{c}{x}$, which I can't see how it relates.

Combinatorically, $\infty^0$ is the number of mappings from the empty set into a set of size "infinity". For a finite and positive $n$, $n^0$ would be the number of mappings from the empty set to a set of cardinality $n$ of which the empty function is the only one. Why does this argument fail? What does it mean to say there is an indeterminate number of mappings, say from the empty set to the set of all natural numbers? I am not familiar with formal set theory (yet) and all I know is that there's something peculiar about infinite cardinalities. So a reference would do in this case if the answer requires knowledge of some theory.

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Your first evaluation is correct; but this is not the only type of "infinity to the 0" limit. How about $$\lim_{x\to\infty} x^{1/\ln (x)}\quad?$$ Again, the base goes to infinity, the exponent goes to $0$. By your "combinatorial argument", the answer "should be" $1$. But it isn't. Proceeding as you do at first, $$\lim_{x\to\infty}x^{1/\ln (x)} = \lim_{x\to\infty}\exp\left(\frac{1}{\ln x}\ln x\right) = \exp\left(\lim_{x\to\infty}\frac{\ln x}{\ln x}\right) = \exp(1) = e.$$ So the answer is $e$, not $1$. The "combinatorial interpretation" doesn't work because even though $x^y$ agrees with the set-theoretic interpretation for integers, it makes no sense for arbitrary real numbers. How are you interpreting $3^{1/3}$ "combinatorially"? The number of maps from a set of size $\frac{1}{3}$ to a set of size $3$?

As to why the answerers mentioned $x\cdot\frac{c}{x}$, the point is that if you have a limit of the form $$\lim f(x)^{g(x)}$$ (I'm omiting what $x$ is going to, because the same argument applies regardless of whether we are taking the limit as $x\to\infty$, as $x\to 3$, as $x\to a$, as $x\to b^{-}$, etc.) where $\lim f(x) = \infty$ and $\lim g(x)=0$, then this is equivalent, exactly as you do at first, with $$\lim f(x)^{g(x)} = \lim\> \exp\left( g(x)\ln(f(x))\right) = \exp\left(\lim g(x)\ln f(x)\right).$$ So it comes down to computing a limit of the form $$\lim g(x)\ln f(x)$$ where $g(x)\to 0$ and $\ln f(x)\to\infty$. This limit can be anything whatsoever, as the examples $x\cdot\frac{c}{x}$ show: set $g(x) = \frac{c}{x}$ (which goes to $0$ as $x\to\infty$) and set $\ln f(x) = x$ (that is, $f(x) = e^x$). Explicitly, $$\lim_{x\to\infty} \left(e^x\right)^{\frac{c}{x}}$$ which is of the form $\infty^0$, is equal to $e^c$, and $e^c$ can be any positive number.

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Thanks this clears up a lot. –  kuch nahi Jun 15 '11 at 5:35

Your L'Hospital's Rule calculation is perfectly correct. If you were doing "analysis" rather than calculus, you might be expected to justify the interchange of $\exp$ and $\lim$, that is, the line that has shape $$\lim_{x \to \infty} (\exp(f(x))=\exp(\lim_{x \to \infty} f(x))$$ but at the calculus level such a justification is not expected. (There are examples where the interchange process leads to the wrong answer, so in principle justification really is needed, but with the "safe" functions we mostly meet in the calculus, problems rarely arise.)

If you have a calculator handy, you might do a spot check of the reasonableness of your answer, by calculating $x^{1/x}$ for some fairly large $x$, like $x=1000$. I get $1.0069317$, which is reasonably close to $1$.

The point of asking you to do the calculation is not only as a check. It is to tell you about the meaning of what you are doing. You are definitely not evaluating something like "$\infty^0$," whatever that might mean. (Actually, it really does not mean anything.)

What you are doing is finding out about the behaviour of the function $x^{1/x}$ when $x$ gets very large, that's all, no mysterious $\infty^0$.

Informal reasoning about the so-called "infinity" is very unreliable. We have moderately good intuition about finite sets. With extreme care, this intuition can be extended to infinite sets. But in any case, we are not dealing with infinite sets. We are dealing with the behaviour of a perfectly concrete function $f(x)$ as $x$ becomes very large.

In particular, the "combinatorial" argument that you give is very imprecise, and it will reliably give you the right answer only when the answer is obvious. It will not deal reliably with any of the so-called indeterminate forms.

It is too bad that the traditional language to describe indeterminate forms uses mnemonics like $\frac{0}{0}$, or $\frac{\infty}{\infty}$, or $\infty^0$. You know very well that we cannot divide $0$ by $0$. Similarly, we cannot meaningfully raise $\infty$, whatever that may mean, to the $0$-th power.

Thinking in terms of infinite sets is irrelevant, and in any case totally unhelpful. Your calculation involved some sort of empty function. Let's look instead at the form traditionally called "$1^\infty$". You might be tempted to think in terms of functions from an infinite set to a $1$ element set. That would be completely unhelpful. For example, the following is a very important result: $$\lim_{n \to\infty}\left(1+\frac{1}{n}\right)^n=e$$ In no way can this result be thought of in terms of functions from an infinite set to a $1$ element set.

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