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I've read that

$$x \log \left(\frac{l+x}{x} \right)=x \log \frac{l}{x} + O(x^2).$$

I tried to derive this using the usual Taylor series method but kept getting a division by zero. Could anyone explain how this identity is derived?

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  • $\begingroup$ Not sure if it helps, but do note that $\frac{l+x}{x} = \frac{l}{x}+\frac{x}{x} = \frac{l}{x} + 1$. But perhaps your Taylor series method already made use of this? $\endgroup$
    – pjs36
    Dec 17 '15 at 2:34
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$$x \log \left(\frac{l+x}{x} \right)=x \log(l/x) + x \log(1 + x/l) = x \log(l/x) + xO(x) = x \log(l/x) + O(x^2).$$

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