using L'Hospital solve $\lim_{x \to \infty} x - x^{2}\ln(1 + \frac{1}{x})$ I can't get this to $ = \frac{0}{0}$ form so I can use l'Hospital rule
$$\lim_{x \to \infty} x - x^{2}\ln\left(1 + \frac{1}{x}\right)$$
tips?
[EDIT]
$$\lim_{x \to 0} \frac{1}{x} - \frac{\ln(1 + x)}{x^{2}}$$
second term $$\lim_{x \to 0} \frac{\ln(1 + x)}{x^{2}} = 
\lim_{x \to 0}  \frac{\frac{1}{1 + x}}{2x} =
2\lim_{x \to 0}  \frac{x}{x+1} = 0$$
first term
$$ \lim_{x \to 0}  \frac{1}{x} = \infty$$
is this okey?

Letting $y=1/x$ gives

$$\lim_{y\to 0} \frac{y-\ln(1+y)} {y^2}. $$

I think you can see it now! 
