Turn on thread page Beta
    • Thread Starter
    Offline

    10
    ReputationRep:
    I have been trying to work this out. It's confusing because wouldn't it just be 0^0, and anything raised to the power of 0 is 1 but 0 raised to any power is 0?

    I rewrote x^x = e^{x \ln x}, but now I am not really sure what to do as when x \to 0 , \ln x \to -\infty .
    • Community Assistant
    Offline

    18
    ReputationRep:
    Community Assistant
    e^xlnx = e^((lnx)/(1/x))
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by NotNotBatman)
    e^xlnx = e^((lnx)/(1/x))
    I don't see how that helps: ln x/1/x. Now if x-->0 then the numerator will tend to negative infinity and the denominator will tend to positive infinity. Is such a quotient defined?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by FXLander)
    I don't see how that helps: ln x/1/x. Now if x-->0 then the numerator will tend to negative infinity and the denominator will tend to positive infinity. Is such a quotient defined?
    You can apply L’Hopital’s on the quotient
    Offline

    11
    ReputationRep:
    (Original post by FXLander)
    I have been trying to work this out. It's confusing because wouldn't it just be 0^0, and anything raised to the power of 0 is 1 but 0 raised to any power is 0?

    I rewrote x^x = e^{x \ln x}, but now I am not really sure what to do as when x \to 0 , \ln x \to -\infty .
    I think that you could look at f(x)=x \ln x. It's easy to show that this is well-defined, decreasing and negative for 0 < x < \frac{1}{e} so as x \to 0_+, we can consider it as an increasing function with supremum 0 (or alternatively look at its reflection -x \ln(-x)). So what is \displaystyle \lim_{x \to 0} f(x)?
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by atsruser)
    I think that you could look at f(x)=x \ln x. It's easy to show that this is well-defined, decreasing and negative for 0 < x < \frac{1}{e} so as x \to 0_+, we can consider it as an increasing function with supremum 0 (or alternatively look at its reflection -x \ln(-x)). So what is \displaystyle \lim_{x \to 0} f(x)?
    It is 0. How do you best show that it is well-defined, decreasing and negative for 0<x<1/e. I was thinking of finding f'(x) = 1+ ln x, but not really sure how to continue from that.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 4, 2017

University open days

  1. Norwich University of the Arts
    Postgraduate Open Days Postgraduate
    Thu, 19 Jul '18
  2. University of Sunderland
    Postgraduate Open Day Postgraduate
    Thu, 19 Jul '18
  3. Plymouth College of Art
    All MA Programmes Postgraduate
    Thu, 19 Jul '18
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.