Proving $\lim_{x\to \infty}\ln(x)/x$ Can you please check if my proofs are correct?
for $$\lim_{x\to +\infty}\ln(x)=+\infty$$
I used the mean value theorem :
$\ln$ continuous on $[1,x]$
$\ln$ differentiable on $(1,x)$
then there exists a point $c$ in $(1,x)$ such that
$$\ln(x)-\ln(1)=(x-1)\ln'(c)$$
$$\ln(x)=(x-1)\cdot1/c$$
$$\lim_{x\to +\infty} \ln(x)=\lim_{x\to +\infty}(x-1)\cdot1/c=+\infty$$
and to prove
$$\lim_{x\to 0+}x\ln(x)=0$$ 
I used the following
$$\lim_{x\to 0+}x\ln(x)=\lim_{x\to 0}ex\ln(ex)=ex\ln(x)+ex$$
$$ A=x\ln(x) / A=eA+ex$$
$$A=ex/(1-e) $$
therefore
$$\lim_{x\to 0+}x\ln(x)=\lim_{x\to 0+}ex/(1-e)=0$$ 
 A: You say that $$\lim_{x\to +\infty}\ln x=\lim_{x\to +\infty} (x-1)\cdot \frac1c=+\infty.$$ This must be justified, since $c$ depends on the interval $[1,x],$ that is, $c$ depens on $x.$ What if $c=\frac{x-1}{2}?$ Then the limit would be $2.$
You can apply the definition of limit:
$$\forall M>0 \exists N=e^M>0 : x>N \implies \ln x>\ln N=\ln e^M=M,$$ to get that $\lim_{x\to \infty}\ln x=\infty.$
With respect to the second part of the question, you say
$$\lim_{x\to 0+} x\ln x=\lim_{x\to 0} (ex\ln (ex))=\lim_{x\to 0}(exln(x)+ex).$$ Since $\lim_{x\to 0} ex$ the equality holds if and only if $\lim_{x\to 0}x\ln x=0.$ If for example the limit is $l\ne 0$ then you would have $l=el,$ which doesn't hold. That is, you can't use this equality to prove $l=0.$
Using L'Hospital's rule:
$$\lim_{x\to 0^+}x\ln x=\lim_{x\to 0^+}\frac{\ln x}{1/x}=\lim_{x\to 0^+}\frac{1/x}{-1/x^2}=\lim_{x\to 0^+} (-x)=0.$$
A: Because $ln$ is increasing it hase a limit $\ell \in \Bbb R\cup \{+\infty\}$ on $+\infty$. 
On other hand $\ln (e^n)=n\to +\infty$ it follows that $l=+\infty$.
EDIT: for $x\geq 1$: 
$\ln \sqrt x=\dfrac{\sqrt x-1}{c(x)}$ with $c(x)\in [1,\sqrt x]$, hence $\dfrac{\sqrt x-1}{\sqrt x}\leq \dfrac{\sqrt x-1}{c(x)}\leq  \sqrt x-1$ i.e  $\dfrac{\sqrt x-1}{\sqrt x}\leq \ln \sqrt x\leq \sqrt x-1 $  so 
$\dfrac{\sqrt x-1}{x\sqrt x}\leq \dfrac{\ln \sqrt x}{x}\leq \dfrac{\sqrt x-1}{x}$. It follows that $\lim_{+\infty}\dfrac{\ln \sqrt x}{x}=0$ and $\lim_{+\infty}\dfrac{\ln  x}{x}=0$.
For $x\ln x$ take $y=\dfrac{1}{x}$.
