Evaluating limit as h->0 of (e^h-1)/h

  • Context: Undergrad 
  • Thread starter Thread starter autodidude
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary

Discussion Overview

The discussion revolves around evaluating the limit as h approaches 0 of the expression (e^h - 1)/h, particularly in the context of differentiating e^x from first principles. Participants explore various methods to handle the limit, including definitions of e and L'Hôpital's rule.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in manipulating the expression (e^h - 1)/h to evaluate the limit without encountering the indeterminate form 0/0.
  • Another participant suggests using the definition of e, lim_{h → 0} (1 + h)^{1/h} = e, to approximate e^h as 1 + h for small values of h.
  • Several participants mention L'Hôpital's rule as a potential method for evaluating the limit, although one participant cautions that using it may involve circular reasoning since the derivative of e^x is what is being sought.
  • A participant questions whether substituting e^h with 1 + h leads back to the 0/0 form when evaluating the limit.
  • Another participant points out the need to simplify the expression to resolve the 0/0 form, suggesting that lim_{h → 0} (1 + h - 1)/h simplifies to lim_{h → 0} h/h.

Areas of Agreement / Disagreement

Participants express differing views on the appropriateness of using L'Hôpital's rule in this context, with some arguing against it while others support its use. The discussion remains unresolved regarding the best approach to evaluate the limit.

Contextual Notes

Some participants highlight the potential for circular logic when applying L'Hôpital's rule without first establishing the derivative of e^x. The discussion also reflects uncertainty about how to manipulate the expression to avoid the 0/0 indeterminate form.

autodidude
Messages
332
Reaction score
0
I'm trying to differentiate e^x from first principles but I can't find a way to manipulate this expression \frac{e^h-1}{h} so I can evaluate the limit without getting 0/0
 
Physics news on Phys.org
Use the definition of e:
lim_{h \rightarrow 0} \left( 1 + h \right)^{1/h} = e
So for small values of h,
e \approx \left( 1 + h \right)^{1/h}, or
e^h \approx 1 + h.

Replace eh in \frac{e^h-1}{h} with 1 + h and go on from there.
 
Diffy said:
l'hospital's rule works here too :-)

http://en.wikipedia.org/wiki/L'Hôpital's_rule

In order to use L'Hopital's rule, you need to know the derivative of ex, but that's what the OP is trying to prove. To use L'Hopital's rule in this case would be circular logic.
 
Diffy said:
l'hospital's rule works here too :-)

http://en.wikipedia.org/wiki/L'Hôpital's_rule

I'm not sure how. The OP is trying to "differentiate ex from first principles," which I assume means that he/she want to find the derivative of ex by using the limit definition:
\frac{d}{dx}e^x = lim_{h \rightarrow 0} \frac{e^{x+h} - e^x}{h} = ...
I don't think you can use l'hospital's' Rule when we "don't know" the derivative of ex yet.EDIT: Beaten to it by Mute. :wink:
 
Oops my bad. I didn't read.
 
eumyang said:
Use the definition of e:
lim_{h \rightarrow 0} \left( 1 + h \right)^{1/h} = e
So for small values of h,
e \approx \left( 1 + h \right)^{1/h}, or
e^h \approx 1 + h.

Replace eh in \frac{e^h-1}{h} with 1 + h and go on from there.

Don't we get 0/0 again?

Limit h->0

\frac{1+h-1}{h}
 
autodidude said:
Don't we get 0/0 again?

Limit h->0

\frac{1+h-1}{h}
Um, you need to simplify the expression.
lim_{h \rightarrow 0}\frac{1+h-1}{h} = lim_{h \rightarrow 0} \frac{h}{h} = ...
 
^ Ah, thanks a lot...must've had a brain meltdown :p
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 53 ·
2
Replies
53
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 46 ·
2
Replies
46
Views
6K
  • · Replies 7 ·
Replies
7
Views
2K