Find $ \lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right) $ with Taylor I have to calculate some limits and try to solve them in use of taylor.
$$ 
\lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right) 
$$
In taylor pattern I have $x_0$ to put, but there $x_0$ is $\infty$ so I want to replace it with something other
$$ 
y = \frac{1}{x} \\
\lim_{y\to 0^+} \left(\frac{1}{y}-\frac{1}{y^2} \ln (1+y)\right) $$
Let $$ f(y) = \frac{1}{y}-\frac{1}{y^2} \ln (1+y) $$
$$f'(y) = -\frac{1}{y^2} + \left(-\frac{2}{y^3}\ln (1+y) - \frac{y^2}{1+y}\right) $$
but $f'(0)$ does not exists because I have $0$ in denominator.

Proceeding with your substitution, since $\log \left( 1+y\right) =y-\frac{1}{
2}y^{2}+O\left( y^{3}\right) $, we have:
\begin{eqnarray*}
\frac{1}{y}-\frac{1}{y^{2}}\log \left( 1+y\right)  &=&\frac{1}{y}-\frac{1}{
y^{2}}\left( y-\frac{1}{2}y^{2}+O\left( y^{3}\right) \right)   
&=&\frac{1}{2}+O\left( y\right) \overset{y\rightarrow 0}{\longrightarrow }
\frac{1}{2}.
\end{eqnarray*}
