Finding the limit of lim w-->wo ((exp(w)-exp(wo))/(w-wo))^-1

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Discussion Overview

The discussion revolves around evaluating the limit of the expression lim w-->wo ((exp(w)-exp(wo))/(w-wo))^-1. Participants explore the methods for finding this limit, including the application of L'Hopital's Rule, and express curiosity about the derivation of the result.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • Some participants assert that the limit equals e^-wo, but seek clarification on the derivation process.
  • One participant suggests that using a different variable layout (x and y instead of w and wo) might improve clarity.
  • Multiple participants propose using L'Hopital's Rule to evaluate the limit, indicating that the limit can be transformed into a reciprocal of another limit involving the derivative of e^w.
  • There is a reiteration of the limit expression and its transformation into a form that emphasizes the evaluation of the derivative at w=wo.

Areas of Agreement / Disagreement

Participants generally agree on the application of L'Hopital's Rule and the limit's result, but there is no consensus on the clarity of the variable representation or the steps leading to the conclusion.

Contextual Notes

Some participants express uncertainty about the steps involved in the limit evaluation and the implications of using different variable representations.

thegirl
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Hi,

I know that when you take this limit it is equal to e^-wo, but I was just wondering how you got there when taking the limit?

lim w-->wo ((exp(w)-exp(wo))/(w-wo))^-1 = 1/e^wo

w and wo are both two points within the same plane.
 
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there's a better layout of the equation with x and y instead of w and wo
 

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thegirl said:
Hi,

I know that when you take this limit it is equal to e^-wo, but I was just wondering how you got there when taking the limit?

lim w-->wo ((exp(w)-exp(wo))/(w-wo))^-1 = 1/e^wo

w and wo are both two points within the same plane.
You can evaluate the expression inside the outer parenthese using L'Hopital's Rule.

$$\lim_{w \to w_0} \left(\frac{e^w - e^{w_0}}{w - w_0} \right)^{-1}$$
The above is equal to
$$\frac{1}{\lim_{w \to w_0} \frac{e^w - e^{w_0}}{w - w_0}} $$
 
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Omg thank you so much!
 
Mark44 said:
You can evaluate the expression inside the outer parenthese using L'Hopital's Rule.

$$\lim_{w \to w_0} \left(\frac{e^w - e^{w_0}}{w - w_0} \right)^{-1}$$
The above is equal to
$$\frac{1}{\lim_{w \to w_0} \frac{e^w - e^{w_0}}{w - w_0}} $$
The denominator in the last expression is simply [itex]\frac{d}{dw}e^w[/itex] evaluated at [itex]w=w_0[/itex].
 
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