Compute a limit $\lim_{x\to\frac 1{e}}\frac{\ln x}{1+\ln x}$ 
Let $f(x)=|\frac{\ln (|x|)}{1+\ln(|x|)}|\space\space\space\space\space\forall\space\space x\in\mathbb{R}\setminus\{-\frac 1e,0,\frac 1e\}$. I want to define this function.

So first I took the function:
$g(x)=\frac {\ln x}{1+\ln x}\space\space\space\forall\space\space x>0, x\neq\frac 1e$.
$g'(x)=\frac 1{x(1+ \ln x)^2}>0 \space\space\space\space\forall\space\space x>0 ,x\neq\frac 1e.$
so $g$ is strictly increasing and $\lim_{x\to0^+}g(x)=1=\lim_{x\to\infty}g(x).$
Now how do I compute the limits at $x=\frac 1e$ from both sides and why are they different? Also when I do limits of functions example when I had $\frac {\infty}{\infty}$ when $x\to0$ and $x\to\infty$, can I use l'hospital to compute them? And when I don't use L'hospital to compute such limits?
 A: Hint: make a substitution $t = \ln x$.
So calculate:
$$\lim_{x\to\frac 1{e}}\frac{\ln x}{1+\ln x}= \lim_{t\to -1}\frac{t}{1+t}$$
Now this limit clearly doesn't exist. Left limit is $+\infty $ and right $-\infty $. 
A: But $$\ln\left(\frac{1}{e}\right)=-1$$ and your denominator is Zero, you can not use the rules of L'Hospital.
If $x$ tends to Zero we get
$\lim_{x\to 0}\frac{\frac{1}{x}}{\frac{1}{x}}=1$ .The same as $x$ tends to infinity.
A: First: Recall that $\ln(x)$ is monotonically increasing on its domain of $(0, \infty)$.
Next: Note that $\ln(1/e) = \ln(e^{-1}) = -1 \ln(e) = -1$.
When $x < 1/e$ by just a bit, $\ln(x) < -1$ by just a bit. And so approaching your limit of $1/e$ from the left, we find a numerator that is getting close to $-1$, and a denominator that is very close to $0$, but ever so slightly negative; so, the negatives cancel and yield a ratio that approaches $+\infty$.
When $x > 1/e$ by just a bit, $\ln(x) > -1$ by just a bit. And so approaching your limit of $1/e$ from the right, we find a numerator that is (again) getting close to $-1$, and a denominator that is very close to $0$, but ever so slightly positive; so, the $-1$ divided by something approaching $0$ but positive yields a ratio that approaches $-\infty$.
To see a graph of this function, as well as its vertical asymptote at $x = 1/e$, check here.
A: We have that the numerator 


*

*$\ln x \to -1$


and the denominator 


*

*$1+\ln x \to 0$


but


*

*$1+\ln x>0$ for $x\to \frac1e^+$

*$1+\ln x<0$ for $x\to \frac1e^-$
therefore we can conclude that le limit doesn’t exist.
