1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find this limit

  1. May 24, 2013 #1

    utkarshakash

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    [itex]\stackrel{lim}{x→0} \left( \dfrac{1^x+2^x+......n^x}{n} \right) ^{1/x}[/itex]

    2. Relevant equations

    3. The attempt at a solution
    Let the quantity inside the bracket be represented by t.Rewriting

    [itex](1+t-1)^{\frac{1}{t-1}.(t-1).\frac{1}{x}} \\
    e^{(t-1)/x} \\
    \stackrel{lim}{x→0} \left( \dfrac{1^x+2^x+......n^x-n}{nx} \right)[/itex]

    Using L Hospital's Rule
    [itex]\stackrel{lim}{x→0} \left( \dfrac{x(ln1+ln2....ln n)}{n} \right)[/itex]

    Now if I put x=0 I get limit as e^0 = 1. But this is not the correct answer.
     
  2. jcsd
  3. May 24, 2013 #2
    if n= 1 our answer is as 1x = 1.

    Take the ln of the expression. Then we get:

    limx->0 ln((1x+...+nx)/n)/x

    If we plug in x = 0 we get ln(1)/0 = 0/0 so we can use L'hopital's rule getting as d/dx x = 1:

    limx->0 n/(1x+..._nx) * (ln(1)1n+...+ln(n)nx)/n =
    limx->0 (ln(1)1x+...+ln(n)nx)/(1x+...+nx).

    This is continuous at 0 so we can plug in x=0 to get:

    (ln(1) + ... + ln(n))/(1+...+1) = (ln(1) + ... + ln(n))/(n).

    So the original limit is exp((ln(1) + ln(2) ... + ln(n))/n)

    edit: checked with matlab this is right.
     
    Last edited: May 24, 2013
  4. May 24, 2013 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You have a limit of the form
    [tex] \lim_{x \to 0} \left( \frac{f_n(x)}{n}\right)^{1/x},\\
    f_n(x) = \sum_{i=1}^n i^x [/tex]
    Never mind l'Hospital; just look at the form of ##f_n(x)## for small |x|, then use some familiar limit results.
     
  5. May 24, 2013 #4

    utkarshakash

    User Avatar
    Gold Member

    I can't understand this step. Please tell me how you have differentiated the numerator?
     
  6. May 25, 2013 #5
    What's the derivative of ##a^x## w.r.t x, where a is some constant?
     
  7. May 25, 2013 #6
    I just canceled the n's. In latex:

    [itex]\frac{n}{1x+..._nx} \frac{ln(1)1n+...+ln(n)nx}{n}[/itex]
     
  8. May 25, 2013 #7
    I just canceled the n's. In latex:

    [itex]\frac{n}{1^x+...+n^x} \frac{ln(1)1^n+...+ln(n)n^x}{n} = \frac{ln(1)1^n+...+ln(n)n^x}{1^x+...+n^x} [/itex]
     
  9. May 25, 2013 #8
    Approach 1: Take natural logarithm on both sides as, ln y = lim x->0 f(x).. Then you apply L' Hospital's Rule.

    Approach 2: Write the given limit in the form of lim n→0 (1+n)1/n = e and then evaluate accordingly.
     
  10. May 25, 2013 #9

    utkarshakash

    User Avatar
    Gold Member

    x lna
     
  11. May 25, 2013 #10

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    No, it's ln a times a^x.
     
  12. May 25, 2013 #11
    No.

    dextercioby has already pointed out the correct derivative. Redo it.
     
  13. May 25, 2013 #12

    utkarshakash

    User Avatar
    Gold Member

    Thanks for pointing out.
     
  14. May 25, 2013 #13

    utkarshakash

    User Avatar
    Gold Member

    Can you type in the exact command that you used to find limit in matlab? It seems my command is giving me incorrect result.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find this limit
  1. Find the Limit (Replies: 3)

  2. Find the limit (Replies: 15)

  3. Finding the Limit? (Replies: 1)

  4. Find the limit (Replies: 26)

  5. Find this limit? (Replies: 4)

Loading...