Results 1 to 7 of 7
Like Tree1Thanks
  • 1 Post By SlipEternal

Math Help - Find limit of differentiable f(x)

  1. #1
    Newbie
    Joined
    Oct 2013
    From
    DC
    Posts
    6

    Find limit of differentiable f(x)

    Let a function f(x) be differentiable at a point x = a and f'(a) = 1. Find the limit

    Find limit of differentiable f(x)-screen-shot-2013-10-01-1.41.16-pm.png
    Attached Thumbnails Attached Thumbnails Find limit of differentiable f(x)-screen-shot-2013-10-01-1.39.39-pm.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1

    Re: Find limit of differentiable f(x)

    Quote Originally Posted by bobcantor1983 View Post
    Let a function f(x) be differentiable at a point x = a and f'(a) = 1. Find the limit
    Click image for larger version. 

Name:	Screen Shot 2013-10-01 at 1.41.16 PM.png 
Views:	11 
Size:	20.7 KB 
ID:	29360
    n\left[ {\sum\limits_{k = 1}^{100} {\left( {f\left( {a + \frac{k}{n}} \right) - 100f(a)} \right)} } \right] = \sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,845
    Thanks
    715

    Re: Find limit of differentiable f(x)

    Quote Originally Posted by Plato View Post
    n\left[ {\sum\limits_{k = 1}^{100} {\left( {f\left( {a + \frac{k}{n}} \right) - 100f(a)} \right)} } \right] = \sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}
    Well, that is not quite true...

    \sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = n\left[ {\sum\limits_{k = 1}^{100} {\left( {f\left( {a + \frac{k}{n}} \right) - f(a)} \right)} } \right] = n\left[ {\left( \sum\limits_{k = 1}^{100}  {f\left( {a + \frac{k}{n}} \right) \right) - 100f(a)} } \right]

    So, your RHS is correct, but the LHS is not (misplaced parenthesis).
    Last edited by SlipEternal; October 1st 2013 at 10:41 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2013
    From
    Washington, DC
    Posts
    2

    Re: Find limit of differentiable f(x)

    I do not quite understand how you went from n\left[ {\left( \sum\limits_{k = 1}^{100}  {f\left( {a + \frac{k}{n}} \right) \right) - 100f(a)} } \right] to n\left[ {\sum\limits_{k = 1}^{100} {\left( {f\left( {a + \frac{k}{n}} \right) - f(a)} \right)} } \right] and incorporated the 100f(a) term into the summation. Also how does the outlying n disappear?

    Assuming that all of this can be reduced down to \sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} would I now consider?

    \lim_{n\to\infty}\sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}

    So here if you look at individual parts of the series you have: f\left( {a + \tfrac{k}{n}} \right) \equiv f(a) since \frac{k}{n} would approach 0 as n approaches infinity. It would follow similarly for the denominator and would cause the whole series to converge to 0?

    How does the fact that f(x) is differentiable at a point x=a and f'(a) = 1 come into play?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,845
    Thanks
    715

    Re: Find limit of differentiable f(x)

    Quote Originally Posted by hughmistler View Post
    I do not quite understand how you went from n\left[ {\left( \sum\limits_{k = 1}^{100}  {f\left( {a + \frac{k}{n}} \right) \right) - 100f(a)} } \right] to n\left[ {\sum\limits_{k = 1}^{100} {\left( {f\left( {a + \frac{k}{n}} \right) - f(a)} \right)} } \right] and incorporated the 100f(a) term into the summation. Also how does the outlying n disappear?
    Let's write out the summation n\left[ {\sum\limits_{k = 1}^{100} {\left( {f\left( {a + \frac{k}{n}} \right) - f(a)} \right)} } \right]:

    n \left[ \left( f\left(a + \dfrac{1}{n} \right) - f(a) \right) + \left( f\left(a + \dfrac{2}{n} \right) - f(a) \right) + \ldots + \left( f\left(a + \dfrac{100}{n} \right) - f(a) \right) \right]
    So, how many times are you subtracting f(a)? Since there are 100 terms, you are subtracting it 100 times.

    The outlying n does not disappear. When you divide by a fraction, that is the same as multiplying by its reciprocal. So, inside the summation we have:
    \begin{align*}k\left[ \dfrac{f\left( a + \tfrac{k}{n} \right) - f(a)}{\tfrac{k}{n}} \right] & = k \dfrac{n}{k}\left[ f\left( a + \dfrac{k}{n} \right) - f(a) \right] \\ & = n\left[ f\left( a + \dfrac{k}{n} \right) - f(a) \right]\end{align*}
    Now, you can factor n from every term, bringing it outside the summation (you can do this since n does not depend on k).

    Quote Originally Posted by hughmistler View Post
    Assuming that all of this can be reduced down to \sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} would I now consider?

    \lim_{n\to\infty}\sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}

    So here if you look at individual parts of the series you have: f\left( {a + \tfrac{k}{n}} \right) \equiv f(a) since \frac{k}{n} would approach 0 as n approaches infinity. It would follow similarly for the denominator and would cause the whole series to converge to 0?

    How does the fact that f(x) is differentiable at a point x=a and f'(a) = 1 come into play?
    Use your limit laws. The limit of a sum is the sum of the limits, so long as those limits exist (and so long as the sum is finite... infinite sums require more work). So:

    \lim_{n\to\infty}\sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = \sum\limits_{k = 1}^{100} {k\left[ \lim_{n\to\infty} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}

    As n\to \infty, \dfrac{k}{n} \to 0. So, we can rewrite this limit:

    \sum\limits_{k = 1}^{100} {k\left[ \lim_{n\to\infty} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = \sum\limits_{k = 1}^{100} {k\left[ \lim_{\tfrac{k}{n}\to 0} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}

    Next, you asked about how you use f'(a) = 1. Use the definition of the derivative. For a general function f(x), you know f'(a) = ??? (Hint: When you write it out, replace h by \tfrac{k}{n})

    For your question about the series converging to 0, no. You have the numerator and denominator approaching zero. 0/0 is not defined, so you need to simplify the expression before you can take the limit.
    Last edited by SlipEternal; October 1st 2013 at 03:19 PM.
    Thanks from dokrbb
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2013
    From
    DC
    Posts
    6

    Re: Find limit of differentiable f(x)

    Ok I think I have this entirely put together, can you tell me if this looks right?

    \lim_{n\to\infty}n[f(a + \frac{1}{n}) + f(a + \frac{2}{n}) + ... + f(a + \frac{100}{n}) - 100f(a)\right]

     = \lim_{n\to\infty}n\left[(\sum\limits_{k=1}^{100}f(a + \frac{k}{n})) - 100f(a)\right]

     = \lim_{n\to\infty}n\left[\sum\limits_{k=1}^{100}(f(a + \frac{k}{n}) - f(a))\right]

     = \lim_{n\to\infty}\sum\limits_{k=1}^{100}k\frac{n}{  k}\left[f(a + \frac{k}{n}) - f(a))\right]

     = \lim_{n\to\infty}\sum\limits_{k=1}^{100}k\left[\frac{f(a + \frac{k}{n}) - f(a)}{\frac{k}{n}}\right]

    Now, since the limit of a sum is the sum of the limits, we can write:

    \lim_{n\to\infty}\sum\limits_{k = 1}^{100} {k\left[ {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = \sum\limits_{k = 1}^{100} {k\left[ \lim_{n\to\infty} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}

    As n\to \infty, \dfrac{k}{n} \to 0. So, we can rewrite this limit:

    \sum\limits_{k = 1}^{100} {k\left[ \lim_{n\to\infty} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = \sum\limits_{k = 1}^{100} {k\left[ \lim_{\tfrac{k}{n}\to 0} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}

    Note: the definition of a derivative is f'(a) = \lim_{h\to\0} \frac{f(a+h) - f(a)}{h}.

    We have this here in our modified limit: \sum\limits_{k = 1}^{100} {k\left[ \lim_{\tfrac{k}{n}\to 0} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]}.

    So we can write \sum\limits_{k = 1}^{100} {k\left[ \lim_{\tfrac{k}{n}\to 0} {\frac{{f\left( {a + \tfrac{k}{n}} \right) - f(a)}}{{\tfrac{k}{n}}}} \right]} = \sum\limits_{k = 1}^{100}kf'(a) = \sum\limits_{k = 1}^{100}k = 5050.


    Is that straightforward and correct?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,845
    Thanks
    715

    Re: Find limit of differentiable f(x)

    Looks good to me
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: February 11th 2011, 01:00 PM
  2. Differentiable and Find df(x)
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: December 13th 2009, 06:52 PM
  3. Replies: 1
    Last Post: April 11th 2009, 11:35 AM
  4. Sequence of differentiable functions, non-differentiable limit
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: April 3rd 2009, 05:13 AM
  5. find when function is NOT differentiable
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 10th 2008, 06:04 PM

Search Tags


/mathhelpforum @mathhelpforum