- Thread starter
- #1

- Thread starter Lisa91
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- Thread starter
- #1

- Jan 29, 2012

- 661

Yes, it is true. One way to prove it: denote, $f(x)=\ln(1+x) - \dfrac{x}{2+x}$, then $f'(x)=\ldots=\dfrac{x^2+2x+2}{(1+x)(2+x)^2}>0$ for all $x>0$. This means that $f$ is strictly increasing in $(0,+\infty)$. On the other hand,How to prove that for [tex] x>0 [/tex]

[tex] \ln(1+x) > \frac{x}{2+x} [/tex] is true?

$\displaystyle\lim_{x\to 0^+}f(x)=\displaystyle\lim_{x\to 0^+}\left(x+o(x)-\frac{x}{2+x}\right)=0$.