How Do I Apply L'Hopital's Rule to Exponential Derivatives?

Click For Summary
SUMMARY

The discussion centers on applying L'Hôpital's Rule to the limit of the expression (2x - 1)^{1/(x-1)} as x approaches 1. Participants clarify that the correct approach involves taking the natural logarithm of both sides, leading to the limit of ln(2x - 1)/(x - 1). This limit can be evaluated using L'Hôpital's Rule, resulting in the limit approaching e. Confusion arises regarding the expected result, with some participants suggesting e^2 instead of e.

PREREQUISITES
  • Understanding of L'Hôpital's Rule
  • Familiarity with limits in calculus
  • Knowledge of the Exponent Rule for Derivatives
  • Basic logarithmic properties and manipulation
NEXT STEPS
  • Study the application of L'Hôpital's Rule in various limit problems
  • Review the Exponent Rule for Derivatives in calculus
  • Practice taking limits involving logarithmic functions
  • Explore examples of limits leading to exponential results
USEFUL FOR

Students and educators in calculus, particularly those focusing on limits and derivatives, as well as anyone seeking to deepen their understanding of L'Hôpital's Rule and its applications.

KristinaMr
Messages
11
Reaction score
1
Homework Statement
how can I find the derivative of (2x-1)^1/(x-1)...the second part is all in the exponent
Relevant Equations
I really don't know which rule applies in this case..maybe there's a way to rearrange the expression?..any help is appreciated
I encountered this problem is Hopital section...how do I even apply it?
 
Physics news on Phys.org
<br /> (2x - 1)^{1/(x-1)} = \exp\left(\frac{\ln (2x-1)}{x - 1}\right)<br />
 
  • Like
Likes   Reactions: KristinaMr and FactChecker
Do you mean L'Hopital? I don't know how that applies to this derivative either. Regarding the derivative, are you familiar with the Exponent rule for derivatives? (see Exponent Rule for Derivatives )
 
FactChecker said:
Do you mean L'Hopital? I don't know how that applies to this derivative either. Regarding the derivative, are you familiar with the Exponent rule for derivatives? (see Exponent Rule for Derivatives )
thank you for the exponent rule ..I somehow missed it

yea it was a limit of x-> 1 so that the exponent would be 1/0 ..the exercise said to apply L Hopital rule to solve ( by the way the result should be e)
 
KristinaMr said:
Problem Statement: how can I find the derivative of (2x-1)^1/(x-1)...the second part is all in the exponent
Relevant Equations: I really don't know which rule applies in this case..maybe there's a way to rearrange the expression?..any help is appreciated

I encountered this problem is Hopital section...how do I even apply it?
This problem has nothing to do with finding the derivative of that function.

Write the function above as an equation: ##y = (2x - 1)^{1/(x - 1)}##
Take the log (ln) of both sides to get ##\ln y = \frac 1 {x - 1} \ln(2x - 1) = \frac{\ln(2x - 1)}{x - 1}##
Now take the limit as x --> 1 of both sides. At this point the limit on the right side can be evaluated using L'Hopital's Rule.

Your textbook should have and example of this technique. Look at the example, and follow the steps in that example.

KristinaMr said:
by the way the result should be e
I don't think so, not as you have shown the problem. I get a value of ##e^2## for the limit.
 
Mark44 said:
I don't think so, not as you have shown the problem. I get a value of ##e^2## for the limit.
No, the limit is as given in the book, e1.
 
ehild said:
No, the limit is as given in the book, e1.
My mistake -- I can't read my own writing...
 

Similar threads

Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
2K