using L'Hospital solve $\lim_{x \to \infty} x - x^{2}\ln(1 + \frac{1}{x})$
Solution 1:
Letting $y=1/x$ gives
$$\lim_{y\to 0} \frac{y-\ln(1+y)} {y^2}. $$
I think you can see it now!
Letting $y=1/x$ gives
$$\lim_{y\to 0} \frac{y-\ln(1+y)} {y^2}. $$
I think you can see it now!