Limits and exponential: $\lim\limits_{x \to\infty} (x-x^2 \ln(1+1/x))$

Question

Evalulate $\lim\limits_{x \to ∞} (x-x^2 \ln(1+1/x))$

I know the answer is 1/2 and we have to make use of the exponential function but i can't seem to simplify the expression to get the answer, please help

http://math.stackexchange.com/questions/1998703/how-do-i-determine-lim-x-to-infty-leftx-x2-log-left1-1-x-right-r — lab bhattacharjee, Nov 07 '16 at 09:36
Do you known limited development of $\ln{1+x}$ when $x\to 0$? — Martigan, Nov 07 '16 at 09:36

score 1 · Accepted Answer · answered Nov 07 '16 at 09:43

$$\lim_{x\to\infty}\left[x-x^2\log\left(1+\frac1x\right)\right]=\lim_{x\to\infty}\frac{\frac1x-\log\left(1+\frac1x\right)}{\frac1{x^2}}\stackrel{\text{l'Hospital}}=$$

$$=\lim_{x\to\infty}\frac{-\frac1{x^2}+\frac1{x^2}\frac x{1+x}}{-\frac2{x^3}}=-\frac12\lim_{x\to\infty}\left(x-\frac{x^2}{x+1}\right)=-\frac12\lim_{x\to\infty}\frac{-x}{x+1}=\frac12$$

Limits and exponential: $\lim\limits_{x \to\infty} (x-x^2 \ln(1+1/x))$

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