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.

You just have to write the Taylor expansion of $\ln \left( 1 + \frac{1}{x} \right)$ when $x$ tends to $+\infty$ :
$$x - x^2 \ln \left( 1 + \frac{1}{x} \right) = x - x^2 \left( \frac{1}{x} - \frac{1}{2x^2} + o\left( \frac{1}{x^2} \right) \right) = \frac{1}{2} + o(1)$$ 
So the limit is equal to $\frac{1}{2}$.
