## Mean Value Theorem Problem

Given f(n) = (1 - (1/n))n

I calculate that the limit as n -> infinity is 1/e.

Also given that x/(1-x) > -log(1-x) > x with 0<x<1 (I proved this in an earlier part of the question) I want to show that:

1 > (f(60)/f(infinity)) > e-1/59 > 58/59

I have tried using my value for f (infinity) and f(60) = e60*log(59/60) in the original inequality but cannot seem to rearrange it to the last one.

I get e(x/1-x) > 1/1-x > ex but am not sure where to go from there.

Thanks!

 Hey sr3056 and welcome to the forums. Try subtracting 1/59 from both sides (let x = 1)
 Mentor Do you know e^x >= 1+x (with "=" just for x=0)? You can use it to show ex/(1-x) > 1/(1-x) 1/1-x > ex... maybe the derivatives can give you some way to prove this. @chiro: x=1 is not in the range where the equation is well-defined.

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## Mean Value Theorem Problem

 Quote by sr3056 I get e(x/1-x) > 1/1-x > ex but am not sure where to go from there.
Try inverting each term in that line (adjusting the relationships as appropriate), then plugging in a value for x.

 Thank you for your responses. Inverting, I get e(x/x-1) < 1 - x < e-x Letting x = 1/60 I then get e-1/59 < 59/60 < e-1/60 This is starting to look a bit more like it, but I'm not sure how my f(60) and f(infinity) values are going to come in...
 Mentor "f(60)/f(infinity)" = e*(1-1/60)^60 Take the log, and your inequality reads: 0 > 1+60 log(1-1/60) > -1/59 You get the left ">" with "-log(1-x) > x" and the right ">" with "x/(1-x) > -log(1-x)", both with the obvious choice for x.
 Thank you very much