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.
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In my opinion, using the Taylor expansion of the logarithm is the easiest way here. – Daniel Fischer May 06 '15 at 10:27
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4See also http://math.stackexchange.com/questions/914072/finding-the-limit-lim-x-to-inftyx-x2-log11-x-in-a-elementary-way and http://math.stackexchange.com/questions/1998703/how-do-i-determine-lim-x-to-infty-leftx-x2-log-left1-1-x-right-r and http://math.stackexchange.com/questions/2003300/limits-and-exponential – Martin Sleziak Nov 09 '16 at 04:51
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$$\log (1+y) = y - y^2/2 + O(y^3).$$ $$\log (1+1/x) = x^{-1} - x^{-2}/2 + O(x^{-3}).$$ $$x^2 \log (1+1/x) = x - 1/2 + O(x^{-1}).$$ $$x^2 \log (1+1/x) -x = - 1/2 + O(x^{-1}).$$ Done!
Mark Joshi
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2How did you determine that expanding up to 2 terms would be enough? – Ava Skovko May 06 '15 at 12:07
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1well there's only 1 term to cancel so develop 1 further to get the lead thing left. More generally, try 2 and if that doesn't work do 3... – Mark Joshi May 06 '15 at 20:17
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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.
KittyL
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