Evaluating $\lim_{x \to e} \left ( \ln(x)\right )^{1/(x - e)}$ with substitutions I evaluated the limit with two substitutions:
$$\begin{align}
L&:=\lim_{x \to e} \left ( \ln(x)\right )^{1/(x - e)}=\\
&=\begin{bmatrix}
t = x - e\\ 
x = t + e
\end{bmatrix}=\lim_{t \to 0} \left ( \ln(t + e)\right )^{1/t}=\\
&=e^{\displaystyle\lim_{t \to 0} \frac{\ln(\ln(t + e))}{t}}\\
\end{align}$$
Now the exponent becomes:
$$\begin{align}
\lim_{t \to 0} \frac{\ln(\ln(t + e))}{t} &= \begin{bmatrix}
\ln(t + e) = 1 + \omega\\ 
t = e^{1 + \omega} - e
\end{bmatrix} = \lim_{\omega \to 0} \frac{\ln(1+\omega)}{e(e^\omega - 1)} =\\ &=\lim_{\omega \to 0}\frac 1e \cdot \frac{\ln(1+\omega)}{\omega} \cdot \frac{\omega}{e^\omega - 1} = \frac 1e
\end{align}$$
So we conclude that $L=e^{1/e}$.  
My question is: can we always substitute as needed or are there constraints that must be met?
I'm asking this because in class the professor used five blackboards, so I thought that maybe there was something wrong with my approach.
 A: First of all I must thank OP for a very nice approach to solving limit problems using standard limit theorems. However there was a small typo (which I have fixed now) as mentioned in comment from John ZHANG.
The technique of substitution is quite useful in limit problems and normally does not require too much constraints. However to be precise the following theorem holds:
If $\lim\limits_{x \to a}f(x) = L$ and $\lim\limits_{t \to b}g(t) = a$ and $g(t) \neq a$ whenever $t$ is in a sufficiently small neighborhood of $b$ (except possibly at $t = b$) then we have $\lim\limits_{t \to b}f(g(t)) = L$.
(Note: The theorem is easy to prove via the usual $\epsilon-\delta$ approach and I would encourage readers to prove it and convince themselves of its validity.)
The last condition which requires $g(t) \neq a$ is essential because here we are effectively substituting $x = g(t)$ and when $x \to a$ we necessarily want $g(t) = x \neq a$. The rule does not apply when the substitution gives $g(t) = a$ for values of $t$ near $b$.
In the problem and solution mentioned by OP there are two substitutions used:
1) $x \to e$ and we put $x = t + e$. Now when $t \to 0$ we see that $x \to e$ but at the same time we also see that $x \neq e$ when $t \to 0$. So this is a valid substitution.
2) $t \to 0$ and $t = e^{1 + \omega} - e$ or $\omega = \log(t + e) - 1$. Here also we see that as $\omega \to 0$ we have $t \to 0$ and $t \neq 0$ when $\omega \to 0$. Thus this substitution is also valid.
A: It is easy to prove this classic result: suppose $\lim_{x\rightarrow x_{0}} f\left(x\right)=\ell\in\mathbb{R}$; if $g$ is defined on a neighborhood of $\ell$ (except possibly $\ell$) and it is continuous at $\ell$, then it does exists $\lim_{x\rightarrow x_{0}} g\circ f$ and $$\lim_{x\rightarrow x_{0}} g(f(x))=\lim_{t\rightarrow\ell}\,g(t)$$
Similarly, if $\lim_{x\rightarrow x_{0}}f(x)=\pm\infty$, you get the same result on the proviso that $\lim_{t\rightarrow \ell}\,g(t)$ exists (finite or infinite). 
A: Let
$y = \displaystyle \lim_{x\to e} (\ln(x))^{\frac{1}{x-e}}$
$\ln(y) = \displaystyle \lim_{x\to e} \frac{1}{x-e}(\ln(\ln(x)))$
You will notice that if you substitute in $e$ directly, you will get $\frac{0}{0}$
On the RHS $\lim_{x\to e} \frac{\frac{1}{x\ln(x)}}{1} = \frac{1}{e}$
$y = e^\frac{1}{e}$
Bottomline, there is always a substitution you can make,here I showed a standard way to tackle this problem. 
You can make substitutions as long as you change the limits in the limit itself and change everything to the substitution in the problem. 
A standard approach I would see is let $u = x - e \implies x = u + e$ 
As $x\to e$ we see that $u \to 0$ 
$\displaystyle \lim_{u \to 0} \ln(u+e)^{u}$
And then take the natural log approach.
Substitutions are fine.
