limit of $\ln x + (x+1)/x$ as $x$ approaches $o$ I want to establish monotone intervals of function $f:(0, \infty) \rightarrow \mathbb R$, where $f(x)=(x+1)\ln x$ using its first derivative. I proved that the first derivative of $f$ is an injective function and I want  to see if $\lim f'(x)$ as $x$ approaches $0$ multiplied by $\lim f'(x)$ as $x$ approaches $1$ is negative, that would mean the equation f'(x)=0 has at least one solution in $(0,1)$, and because $f'$ is injective there would be exactly one solution in $(0,1)$. Then I would be able to establish monotone intervals of $f$, even without knowing when $f'(x)=0$, I'll just name the solution '$y$'. But first I have to know if it there is only one solution y, and I don't know what is the limit of $\ln x + (x+1)/x$ as $x$ approaches $0$.
 A: For your limit: 
$$\ln(x) + \frac{x+1}{x} = 1 + \frac{1 + x \ln(x)}{x} \\
\lim_{x \to 0^+} x \ln(x) = 0 \\
\Rightarrow \lim_{x \to 0^+} \ln(x) + \frac{x+1}{x} = \lim_{x \to 0^+} 1 + \frac{1}{x} = +\infty.$$
For the stuff about monotonicity:
The function $g(x)=x \ln(x)$ is minimized when $x=e^{-1}$, where it is equal to $-e^{-1}$. You can see this because $g'(x)=\ln(x)+1$, which is zero when $x=e^{-1}$. Taking another derivative, you get $g''(x)=1/x>0$, so the critical point is a local minimum. This minimum is in fact global, because $g''(x)>0$ for all $x$.
Since $e^{-1}<1$, the numerator is positive for all $x>0$, so your function is positive for every $x>0$. 
A: The derivative of $f(x)=(x+1)\ln x$ is
$$
f'(x)=\ln x+\frac{x+1}{x}=\ln x+ 1 +\frac{1}{x}
$$
The second derivative is
$$
f''(x)=\frac{1}{x}-\frac{1}{x^2}=\frac{x-1}{x^2}
$$
which shows $f'(x)$ has a minimum at $1$; since
$$
f'(1)=2>0
$$
we can say that $f'(x)>0$ for all $x$. Therefore $f$ is strictly increasing.
Note that $f'$ is not injective. Indeed
$$
\lim_{x\to0}f'(x)=\lim_{x\to0}\frac{x\ln x+1}{x}+1=\infty
$$
and
$$
\lim_{x\to\infty}f'(x)=\infty
$$


The same with GeoGebra

A: Hint: Letting $t=\dfrac1x$ we have $~\displaystyle\lim_{x\to0}\bigg(\ln x+\frac1x\bigg)~=~\lim_{t\to\infty}(t-\ln t),~$ which is trivial.
