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Math.

MIT 18.01 Single Variable Calculus (OCW)
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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
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Other answers:

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But then, you still haven't considered lnx...
why do you need to prove ... ln(2) > 1/2 ?
It's not prove.. but.. we have to consider lnx..
why??
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
x/(1+x) < ln(1+x) < x ... perhaps we can use the same argument as we used above.
\[ (x+1)^2 > 1+x > 1 \text{ for all x > 0} \\ {1 \over (1+x)^2} < {1 \over 1 +x} < 1 \text{ for all x > 0} \\ \int_0^x {1 \over (1+x)^2} \;dx < \int_0^x {1 \over 1 +x}\;dx < \int_0^x1 \; dx \\ {x \over 1+x} < \ln(1+x) < x \]
\[ \ln (x) \text{ in } [1, 1+t], t>0 \] perhaps this would mean show that \( \ln(1)<{t \over 1 +t} < \ln(1+t)\)
MVT http://upload.wikimedia.org/wikipedia/commons/thumb/e/ee/Mvt2.svg/300px-Mvt2.svg.png
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f(b) - f(a) / (b-a) ?
yep!!
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OMG! My super smart brother!!
lol ... not that smart. Just copied half from your prev answer.

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