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I am trying to use the basic laws of limits (ie. L'Hospital's rule) to find a solution to:

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

I know that as a general rule:

$$\lim_{x\to c} \left(f(x) - g(x)\right) = \lim_{x\to c} f(x) - \lim_{x\to c} g(x)$$

but since $\lim_{x\to \infty} x$ diverges, I'm not sure how to approach this problem. I assume this property I described above only works for two convergent expressions.

Any guidance to point me in the right direction would be greatly appreciated!

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    $\begingroup$ You are right, you can only distribute the minus sign over the limit if one of the limits of what you are subtracting converges. In your case, both limits diverge after you distribute the minus sign so you can't do that. $\endgroup$ – user2566092 Mar 30 '15 at 17:36
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Setting $1/x=h$

$$\lim_{h\to0}\frac{h-\ln(1+h)}{h^2}$$

Method $\#1:$

Use Series Expansion , $\ln(1+h)=h-\dfrac{h^2}2+O(h^3)$

Method $\#2:$

As the expression is of the form $\dfrac00,$ safely use L'Hospital's Rule

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An alternative approach. $$x-x^2\log\frac{1+x}{x} = \int_{x}^{x+1}\left(y-\frac{1}{2}-\frac{x^2}{y}\right)\,dy=\int_{x}^{x+1}\frac{2y^2-y-2x^2}{2y}\,dy$$ and: $$\int_{x}^{x+1}\frac{2y^2-y-2x^2}{2y}\,dy = -\frac{1}{2}+\int_{0}^{1}\frac{t+2x}{t+x}t\,dt$$ hence by the dominated convergence theorem: $$\lim_{x\to +\infty}\left(x-x^2\log\frac{1+x}{x}\right)=-\frac{1}{2}+\int_{0}^{1}2t\,dt = \color{red}{\frac{1}{2}}.$$

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    $\begingroup$ This is really nice. +1. Seen this approach for the first time. $\endgroup$ – Paramanand Singh Mar 30 '15 at 18:08
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Setting $y=\frac{1}{x}$ we can rewrite the limit as

$$\lim_{y\to 0}\frac{1}{y}\left[1-\frac{\log{(1+y)}}{y}\right]$$

Now $\frac{\log(1+y)}{y}=\frac{y-\frac{y^2}{2}+o(y^2)}{y}=1-\frac{y}{2}+o(y)$. And this leads to

$$\lim_{y\to 0}\frac{1}{y}\left[1-\frac{\log{(1+y)}}{y}\right]=\frac{1}{2}$$

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