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Prove using mean-value theorem that $x/(1+x^2)<\arctan x<x$ for $x>0$

I got the first part but how do I prove $\arctan x< x$ using the MVT?

The first part was done easily by applying MVT on $\arctan x$, should I use $\arctan x-x$ for the second part? Thanks!

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closed as off-topic by 6005, Claude Leibovici, user91500, Alex M., Tom-Tom Nov 30 '15 at 13:53

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I think you have it. One application of the MVT gives you both parts:

Let $x>0$. Applying the Mean Value Theorem to $f(x)=\arctan x$ on the interval $[0,x]$ gives a number $c$ with $0<c<x$ such that $$ {\arctan x-\arctan 0\over x-0}={1\over 1+c^2} $$ Rearranging the above gives: $$ \arctan x={x\over 1+c^2} . $$ for some $c$ between $0$ and $x$.

Since $x>c$ and $x\gt0$, we have: $${x\over 1+x^2}\lt{x\over 1+c^2}<x;$$ whence $$ {x\over 1+x^2}\lt\arctan x\lt x . $$

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  • $\begingroup$ thnx alot! perfect ans! $\endgroup$ – tirmizi Dec 31 '11 at 5:00

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