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 $0$.

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 $x>0$. 
