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

• How do I determine $\lim_{x\to\infty} \left[x – x^{2} \log\left(1 + 1/x\right)\right]$?

#### Solutions Collecting From Web of "Limit $\lim_{x\to\infty} x-x^{2}\ln\left(1+\frac{1}{x}\right)$"

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

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.