Limit $\lim_{x\to\infty} x-x^{2}\ln\left(1+\frac{1}{x}\right)$

Question

Give me please a hint, how to find

$$\underset{x\rightarrow\infty}{\lim}x-x^{2}\ln\left(1+\frac{1}{x}\right)$$

I tried to use substitution $h = \frac{1}{x}$ and, then, apply L'Hopital's rule, but it went worse.

Answer 1

$$\log (1+y) = y - y^2/2 + O(y^3).$$ $$\log (1+1/x) = x^{-1} - x^{-2}/2 + O(x^{-3}).$$ $$x^2 \log (1+1/x) = x - 1/2 + O(x^{-1}).$$ $$x^2 \log (1+1/x) -x = - 1/2 + O(x^{-1}).$$ Done!

Answer 2

Sorry I was not right just now.

You can do substitution as you mentioned.

$$\lim_{h\rightarrow 0}\left(\frac{1}{h}-\frac{1}{h^2}\ln{(1+h)}\right)\\ =\lim_{h\rightarrow 0}\frac{h-\ln{(1+h)}}{h^2}\\ =\lim_{h\rightarrow 0}\frac{1-\frac{1}{1+h}}{2h}\\ =\lim_{h\rightarrow 0}\frac{h}{2h(1+h)}$$

And you can continue from here. From line 2 to 3 I used L'Hospital's rule.