What is the limit of (1-tanh(x))/e^(-2x)?

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The limit of (1-tanh(x))/e^(-2x) as x approaches infinity is confirmed to exist and equals 2, despite initial confusion regarding its behavior. The expression simplifies to e^(2x) - (e^(3x)-e^(x))/(e^x+e^(-x)), which leads to oscillatory behavior when plotted in Maple. The discussion highlights the importance of accurate plotting and verification, as initial results suggested divergence. Ultimately, the correct interpretation reveals that the limit converges to 2.

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examine the limit of x→∞ of:

(1-tanh(x))/e^(-2x) = (1- (e^x-e^(-x))/(e^x+e^(-x)))/e^(-2x)

Rearranging a bit we get:

e^(2x) - (e^(3x)-e^(x))/(e^x+e^(-x))

Now plotting it in maple it seems to behave very badly, it oscillates up and down. Problem is prooving that there indeed exists no limit.

Intuitively when x gets big the e^(3x)/(e^(x)+e^(-x)) term should approach e^(2x) - but for some reason IT DOES NOT. What is going on with this crazy function and does anyone have ideas how to proove that for a given a i can never find a delta such that lf-al ≤ δ etc etc.
 
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According to WolframAlpha, the limit exists and it is 2. Are you sure you plotted the right function in maple?
 
yes I plotted:

(1-tanh(x))/(e^(-2x)) are u sure U plotted the right one? :) Else maybe I will resort to wolfram alpha. It seems that maple messes up sometimes. For instance it showed this as diverging too:

e^(2x)-e^(3x)/e^x and surely that can't be right?

edit: nvm you got me. I made a silly mistake - stupid me :(
 

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