How Does the Limit of cosh and sinh Approach 1?

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Discussion Overview

The discussion revolves around the limits of the hyperbolic functions sinh and cosh as their argument approaches infinity, particularly in the context of a physical expression involving temperature and magnetic fields. Participants explore how these limits relate to logarithmic expressions and the behavior of these functions as certain parameters approach zero.

Discussion Character

  • Technical explanation, Mathematical reasoning, Conceptual clarification

Main Points Raised

  • One participant questions how the limit of ln(infinity) relates to the expression uB/KbT and seeks clarification on how sinh and cosh approach 1.
  • Another participant asserts that both sinh(x) and cosh(x) approach e^x as x approaches infinity, suggesting that the 1 term becomes negligible compared to the cosh term.
  • A participant reiterates the previous point about the limits of cosh and e^x, indicating that as x approaches infinity, both functions approach infinity, leading to the conclusion that lim(x->infinity) cosh(x) equals lim(x->infinity)e^x.
  • One participant provides a mathematical expression showing that as T approaches 0, ln(1 + 2cosh(μB/kBT)) approximates to μB/kBT.
  • A participant explains that in the expression for cosh(x), the e^(-x) term becomes negligible compared to the e^x term as x approaches infinity.

Areas of Agreement / Disagreement

Participants express similar views regarding the behavior of cosh and sinh as their arguments approach infinity, but the discussion remains unresolved regarding the specific limit of ln(infinity) and its relation to the other expressions mentioned.

Contextual Notes

There are unresolved assumptions regarding the definitions and limits involved in the expressions, particularly concerning the transition from logarithmic to hyperbolic functions and the implications of temperature approaching zero.

thegirl
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Hi I was wondering how you get this when taking the limit of T going to 0
Screen Shot 2016-03-30 at 16.44.47.png

From this expression of S:
Screen Shot 2016-03-30 at 16.49.30.png


Please help I don't see how ln infinity goes to uB/KbT (used u to represent the greek letter. And how does the other expression of sinh and cosh approach 1?
 
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Both sinh(x) and cosh(x) approach e^x as x->infinity. Do you see why? So the 1 term becomes negligible compared to the cosh term. Does this help?
 
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phyzguy said:
Both sinh(x) and cosh(x) approach e^x as x->infinity. Do you see why? So the 1 term becomes negligible compared to the cosh term. Does this help?
is that because when x aproaches infinity cosh(x) = infinity and as e^x also approaches infinity when x approaches infinity it can be said that lim(x-> infinity) cosh(x) = (lim x-> infinity)e^x? Also, thanks for replying!
 
As ##T\rightarrow 0## we have that ##\ln{\left(1+2\cosh{\left(\frac{\mu B}{k_{B}T}\right)}\right)}\sim \ln{\left(1+ e^{\left(\frac{\mu B}{k_{B}T}\right)}\right)}\sim \frac{\mu B}{k_{B}T}## ...
 
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thegirl said:
is that because when x aproaches infinity cosh(x) = infinity and as e^x also approaches infinity when x approaches infinity it can be said that lim(x-> infinity) cosh(x) = (lim x-> infinity)e^x? Also, thanks for replying!

cosh(x) = \frac{e^x + e^{-x}}{2}

As x->infinity the e^(-x) term approaches zero and becomes negligible compared to the e^x term.
 

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