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experimentX
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it is given to prove that x/(1+x) < ln(1+x) < x in the interval [1,1+t] t>0

experimentX
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dw:1350718620013:dw

experimentX
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dw:1350718731005:dw

experimentX
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dw:1350718795042:dw

Callisto
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But then, you still haven't considered lnx...

experimentX
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why do you need to prove ... ln(2) > 1/2 ?

Callisto
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It's not prove.. but.. we have to consider lnx..

experimentX
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why??

experimentX
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By considering ln x in [1; 1 + t] (t > 0),
ln x in [1; 1 + t] (t > 0),
minimum bound : ln(x) = 1 => x = e

experimentX
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x/(1+x) < ln(1+x) < x ... perhaps we can use the same argument as we used above.

experimentX
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\[
(x+1)^2 > 1+x > 1 \text{ for all x > 0} \\
{1 \over (1+x)^2} < {1 \over 1 +x} < 1 \text{ for all x > 0} \\
\int_0^x {1 \over (1+x)^2} \;dx < \int_0^x {1 \over 1 +x}\;dx < \int_0^x1 \; dx \\
{x \over 1+x} < \ln(1+x) < x
\]

experimentX
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\[ \ln (x) \text{ in } [1, 1+t], t>0 \]
perhaps this would mean
show that \( \ln(1)<{t \over 1 +t} < \ln(1+t)\)


experimentX
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dw:1350722628100:dw

experimentX
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dw:1350722670829:dw

experimentX
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dw:1350722732492:dw

experimentX
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dw:1350722797470:dw

experimentX
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dw:1350722973766:dwdw:1350723003576:dw

experimentX
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dw:1350723085914:dw

Callisto
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f(b)  f(a) / (ba) ?

experimentX
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yep!!

experimentX
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dw:1350724369177:dw

Callisto
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OMG! My super smart brother!!

experimentX
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lol ... not that smart. Just copied half from your prev answer.