# Prove $\lim_{x\to\infty}x^{\ln(x)} = \infty$

I am trying to prove

$$\lim_{x\to\infty}x^{\ln(x)} = \infty$$

I am going to break this into two methods: one my professor mentioned and my method (which is where the question lies - skip ahead if you must!). Note that this is not homework, but simply an exercise my professor decided to do during notes the other day. The problem is taken from Stewart 7e Calculus (#70a in section 6.3 if you want to bust out your (e)-book).

Method 1:

Recall $\ln(e^x) = x, e^{\ln(x)} = x$. Thus we can write the original limit as $$\lim_{x\to\infty}\left(e^{\ln(x)}\right)^{\ln(x)} = \lim_{x\to\infty}e^{\left(\ln (x)\right)^2}$$

He then let $u = \ln(x)$. As $x\to\infty$, then $u=\ln(x) \to\infty$. As $u\to\infty, v = u^2 \to\infty$. Also, as $v\to\infty, e^v \to\infty$. So, as $x\to\infty, e^{\left(\ln (x)\right)^2} \to\infty$. Thus it is sufficient to say $$\lim_{x\to\infty}x^{\ln(x)} = \infty \ \ \ \ \ \ \ \mathrm{Q.E.D.}$$

Method 2 (my attempt):

Let $t = \ln x$. As $x\to\infty, t\to\infty$ because the $\ln$ function is strictly increasing.

$$\lim_{x\to\infty}x^{\ln(x)} \equiv \lim_{t\to\infty}x^t \tag{1}$$

Does the last statement of line (1) make sense mathematically though since the limit is with the variable $t$, yet the argument contains an $x$ still?

Since line (1) may not be formally correct, I decided to try to write $x$ in terms of $t$. Recall that I made the substitution $t = \ln x \implies e^t = e^{\ln x} = x$. Thus I rewrote the limit as

$$\lim_{t\to\infty}\left(e^t\right)^t$$ which diverges to $\infty$ for sufficiently large values of $t$.

As an added bonus, are there any other 'simple' proofs for this limit?

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I deleted my answer since it was just (1) above. –  DonAntonio Sep 5 '12 at 3:53
The simple proof is that this is a determinate limit form: $(+\infty)^{+\infty} = +\infty$ –  Hurkyl Sep 5 '12 at 4:08
For method 2, would it be possible to use the fact that $\ln{(x)} > 0$ for $x > 1$, and that $\displaystyle \lim_{x \to \infty} x^n = \infty$ if $n > 0$? –  Thomas Sep 5 '12 at 6:33

Your statement (1) does not really make sense for the reason you cite. Also note that "converges to $\infty$" is generally not correct; instead one usually says "diverges to $\infty$".

Simple proof: Observe that for $x\geq e$, we have $x^{\ln x}>x$. Thus $\lim\limits_{x\to\infty} x^{\ln x}\geq \lim\limits_{x\to\infty} x$, which is clearly $\infty$.

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Nice proof! Regarding converges vs. diverges, I just caught myself when re-reading through my post - about the same time your edit came in. Thanks. –  Joe Sep 5 '12 at 3:54
@Joe: I confess that I prefer "converges to $+\infty$" despite it being uncommon phrasing. But it sort of depends on your point of view: are you thinking of the function converging to a particular point of the extended real line, or are you thinking of a specific way in which the limit can fail to converge to a point of the (ordinary) real line? The latter is how things are usually introduced, but IMO the former is really the right way to think about it. –  Hurkyl Sep 5 '12 at 4:12
@Hurkyl My line of thought was your former viewpoint. I've seen it both ways in texts and from other professors. Diverge is definitely the majority in this argument though. I never truly learned the whole "Why?" for the differences though - beyond intuitively, I suppose. If you want to elaborate, that would be nice! –  Joe Sep 5 '12 at 4:16
@Hurkyl I agree that converges to $\infty$ is fine if you are working consistently in some compactification of $\mathbb R$. But usually one is not. –  Alex Becker Sep 5 '12 at 4:33

Your $(1)$ isn’t helpful, because it loses (or at least obscures) the relationship between the base and the exponent of the expression, and that relationship is crucial. Carrying the substitution through fully, as you did when you wrote

$$\lim_{t\to\infty}\left(e^t\right)^t\;,$$

is just fine and does indeed show that $\lim_{x\to\infty}x^{\ln x}=\infty$.

Note, though, that’s it’s not correct to say that $\lim_{t\to\infty}\left(e^t\right)^t$ ‘converges to infinity for sufficiently large values of $t$. It isn’t the limit that’s converging: it’s the expression $\left(e^t\right)^t$. But it still wouldn’t have been correct if you’d said that $\left(e^t\right)^t$ converges to infinity for sufficiently large values of $t$: it converges to infinity as $t$ increases without bound, or as $t$ tends to infinity. When we say that statement $P(t)$ about $t$ is true for sufficiently large values of $t$, we mean that there is some number $t_0$ such that $P(t)$ is true whenever $t\ge t_0$. But ‘$\left(e^t\right)^t$ converges to infinity’ isn’t a statement about specific real numbers $t$: it’s a statement about the function $f(t)=\left(e^t\right)^t$.

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Thanks for the info about my misuse of "sufficiently large values of t". I'll note that for the future! –  Joe Sep 5 '12 at 4:03

How about taking log of your expression? Then we have that $$\lim_{x\to\infty}\ln^2(x)\longrightarrow \infty < \lim_{x\to\infty}x^{\ln(x)}$$ and the conclusion follows.

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