calculus - Prove $lim_{xtoinfty}x^{ln(x)} = infty$

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?

Answer

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 $\infty$.