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Leibniz Limits

  1. Jan 15, 2006 #1

    Is this identity true?

    Identity:
    [tex]\frac{d}{dx} x^n = \lim_{h \rightarrow 0} \frac{(x + h)^n - x^n}{h} = nx^{n-1}[/tex]

     
    Last edited: Jan 15, 2006
  2. jcsd
  3. Jan 15, 2006 #2

    benorin

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    Sure it's true, for non-negative integer n. Even for all [itex]n\neq -1[/itex].
     
  4. Jan 15, 2006 #3

    HallsofIvy

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    ?? Even for n= -1 (as long as x is not 0) and, indeed, for n any complex number (again, as long as x is in the domain). That's why that formula is typically the first derivative formula one learns in calculus!
     
  5. Jan 19, 2006 #4


    Is it possible to solve the limit part of this equation in a classical way to produce the solution?

    [tex]\frac{d}{dx} (\ln x) = \lim_{h \rightarrow 0} \frac{\ln(x + h) - \ln x}{h}[/tex]
     
    Last edited: Jan 19, 2006
  6. Jan 19, 2006 #5

    VietDao29

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    Yes, it is.
    Lemma:
    [tex]\lim_{x \rightarrow 0} \frac{\ln(1 + x)}{x} = 1[/tex]
    Proof:
    [tex]\lim_{x \rightarrow 0} \frac{\ln(1 + x)}{x} = \lim_{x \rightarrow 0} \frac{1}{x} \ \ln(1 + x) = \lim_{x \rightarrow 0} \ln \left[ (1 + x) ^ \frac{1}{x} \right] = \ln \lim_{x \rightarrow 0} \left[ (1 + x) ^ \frac{1}{x} \right] = \ln \lim_{y \rightarrow \infty} \left[ \left( 1 + \frac{1}{y} \right) ^ y \right] = \ln e = 1[/tex].
    (y = 1 / x)
    ----------------------
    [tex]\lim_{h \rightarrow 0} \frac{\ln(x + h) - \ln x}{h} = \lim_{h \rightarrow 0} \frac{\ln \left( \frac{x + h}{x} \right)}{h} = \lim_{h \rightarrow 0} \frac{\ln \left( 1 + \frac{h}{x} \right)}{h} = \lim_{h \rightarrow 0} \frac{1}{x} \ \frac{\ln \left( 1 + \frac{h}{x} \right)}{ \left( \frac{h}{x} \right)} = \lim_{k \rightarrow 0} \frac{1}{x} \ \frac{\ln \left( 1 + k \right)}{k} = \frac{1}{x}[/tex] (Q.E.D)
    k = h / x
    ----------------------
    I think you can find this in your calculus book... :rolleyes:
     
  7. Jan 19, 2006 #6

    benorin

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    Sounds like the constraint for the integral analog of that formula, doesn't it? :uhh: I should sleep more...
     
  8. Jan 19, 2006 #7

    mathwonk

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    this is a nice argument vietdao29. however one can prove anything if you assume enough background. In my opinion the things you have used in your proof, are more difficult to prove than the result you have proved using them.

    For instance the existence alone of the log function and its continutiy and various properties, are all usually proved using the derivative formula. Of course they can be proved otherwise, but it is more difficult than the argument you are using them for.

    I would ask how do you prove all those difficult properties you are using without knowing the derivative in advance?

    One particular example, the limit calculation of e, is itself proved in my calculus book using the derivative formula for ln(x). Thus using my calculus book's argument to complete yours, would render it circular.

    Dieudonne has a nice developoment of the log function and its continuity and multiplicativity properties without derivatives, defining the exponential as its inverse. It follows that the exponential is continuous hence integrable.

    He then shows that a^x can be written as a constant times the difference of two values of its own integral, and since the integral of a continuous function is differentiable, the differentiability and derivative formula for the exponential follows, and hence also that for the log.

    still he does not derive the limit (I + (1/x))^x --> , as x -->infinity, directly.

    this is usually derived by l'hopitals rule, i.e. exactly the reverse of your aregument. thus i ask how you prove that limit without knowing any derivative formulas for ln or exp?

    i.e. that limit you are assuming is essentialy equivalent to the one you are proving, so you have not made any progress unless you show how to prove one of them without using the other.
     
    Last edited: Jan 19, 2006
  9. Jan 20, 2006 #8
    [tex]\frac{d}{dx} (\ln x) = \lim_{h \rightarrow 0} \frac{\ln(x + h) - \ln x}{h}[/tex]
     
  10. Jan 20, 2006 #9

    mathwonk

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    huh?

    i am asking for a proof of lim x-->infinity [1 + 1/x]^x = e, which

    does not use the limit:

    lim h-->0 [ln(x+h)-ln(x)]/h = 1/x.
     
  11. Jan 21, 2006 #10

    VietDao29

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    Uhmm, my book define e to be:
    [tex]e = \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right) ^ n, \ n \in \mathbb{N}[/tex]
    It can be shown that e is bounded using Binomial theorem.
    Letting
    [tex]u_n = \left( 1 + \frac{1}{n} \right) ^ n[/tex] we can show that {un} is increasing and is bounded, lower bounded by 2, and upper bounded by 3.
    From there, it can be shown that:
    [tex]e = \lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x, \ x \in \mathbb{R}[/tex] by the Squeeze theorem.
    Since we have:
    If [tex]\lim_{x \rightarrow \alpha} f(x) = L[/tex] then [tex]\lim_{x \rightarrow \alpha} f(x) ^ n = L ^ n[/tex] (n is some constant).
    We then can show that:
    [tex]e ^ x = \lim_{k \rightarrow \infty} \left( 1 + \frac{x}{k} \right) ^ k, \ k \in \mathbb{R}[/tex]
    -------------------------
    From here, my book assumes that ex is continuous, and is increasing (since e > 1). (?)
    They state that based on the fact that if a > 1 Then ax is increasing. However, they don't prove that fact. They say that it's generally accepted!!! (?)
    I think I need to consult my maths teacher about this.
    -------------------------
    So I think about some other way:
    We can prove that:
    (xa)' = axa - 1, for all a in the reals.
    We can also prove the chain rule using limit.
    That means:
    [tex](e ^ x)' = \lim_{k \rightarrow \infty} \left( \left( 1 + \frac{x}{k} \right) ^ k \right)' = \lim_{k \rightarrow \infty} \left( 1 + \frac{x}{k} \right) ^ {k - 1} = e ^ x, \ k \in \mathbb{R}[/tex]
    Using the fact that e0 = 1, and (ex)' = ex, we can show that:
    [tex]e ^ x = 1 + \sum_{n = 1} ^ \infty \frac{x ^ n}{n!} = \sum_{n = 0} ^ \infty \frac{x ^ n}{n!}[/tex]
    From here, we can say that ex is continuous, increasing, and can prove some of its properties like:
    ea + b = ea eb, ...
    Since ex is increasing, it must have an inverse function, known as ln(x) (whose graph is the reflection of the graph ex across the line y = x). So ln(x) is continuous (since ex is continuous).

    From here, I think I can show that (ln(x))' = 1 / x using the derivatives for inverse function. But since my book generally accept things, so their way is much longer!!!!!
    Using : ln(x) = a <=> ea = x. We can show that: eln a = a. From here, we can prove all proterties for the log function, like:
    [tex]\ln(x) - \ln(a) = \ln \left( \frac{x}{a} \right)[/tex]
    [tex]\ln(x) + \ln(a) = \ln (xa)[/tex]
    [tex]\ln(x ^ a) = a \ln (x)[/tex], ...
    Is there any fraud in my reasoning?
    Am I using good terminology? (English is not my mother tongue :tongue2:)
    -------------------------
    It would be nice if you can show me your book's definition of e. And how can they prove some log and exp's properties...
    And may I know the name of the book?
     
    Last edited: Jan 21, 2006
  12. Jan 21, 2006 #11

    matt grime

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    agreed for integer n
    no, you can't say that. how did you go from integer n to k in R?
    how can you even prove that? what is x^a if a is not an integer or a rational?

    exp(x) is the uique solution to f'=f f(0)=1, it exists and is well defined, it has powerseries we know and love.

    In anycase, e=1+1+1/2!+1/3!+...

    and you haven't proved that is equal to the limit of (1+1/n)^n.
     
    Last edited: Jan 21, 2006
  13. Jan 21, 2006 #12

    benorin

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    Let us be rigorous! Tell us what, then, is known at the point of the question that we avoid avoid circular reasoning and knowledge of theorems more advanced that that which is to be proved: how is the [itex]\ln x[/itex] defined? if it is as the inverse of [itex]e^x,[/itex] how was that defined? do we admitt such theorems as the product, quotient chain rules? the binomial theorem?

    Are you looking for an [tex]\epsilon -\delta[/tex] proof? be specific, please.
     
  14. Jan 21, 2006 #13

    benorin

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    It's right here.

    It's right here.
     
  15. Jan 21, 2006 #14

    matt grime

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    Oh, and log(x) is (equivalent to being defined as) the integral from 1 to x of 1/t dt. You can prove everything you want to about logs from that definition, by the way, ie that log(xy)=log(x)+log(y) and that log(x^r)=rlog(x), in particular that log(1/x)=-log(x)
     
    Last edited: Jan 21, 2006
  16. Jan 21, 2006 #15

    HallsofIvy

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    And, perhaps most importantly, that its inverse is an exponential.
     
  17. Jan 21, 2006 #16

    In order to proceed with the solution to the limit definition for the natural logarithm (ln) derivative, we must first prove the limit definition for base (e).

    [tex]f(x) = \ln x[/tex]

    [tex]f'(1) = \lim_{h \rightarrow 0} \frac{f(1 + h) - f(1)}{h} = \lim_{x \rightarrow 0} \frac{f(1 + x) - f(1)}{x}[/tex]

    [tex]= \lim_{x \rightarrow 0} \frac{\ln(1 + x) - \ln 1}{x} = \lim_{x \rightarrow 0} \frac{1}{x} \ln(1 + x)[/tex]

    [tex]= \lim_{x \rightarrow 0} \ln [(1 + x)^{\frac{1}{x}}][/tex]

    [tex]= \lim_{x \rightarrow 0} \ln [(1 + x)^{\frac{1}{x}}] = \ln [\lim_{x \rightarrow 0} (1 + x)^{\frac{1}{x}}] = \ln e = 1[/tex]

    [tex]f'(1) = 1[/tex]

    [tex]\text{Theorem:}[/tex]
    [tex]\text{If} \; f \; \text{is continuous at b and} \; \lim_{x \rightarrow a} g(x) = b \; \text{then} \; \lim_{x \rightarrow a} f(g(x)) = f(b)[/tex]

    [tex]\lim_{x \rightarrow a} f(g(x)) = f (\lim_{x \rightarrow a} g(x))[/tex]

    [tex]e = e^1 = e^{\lim_{x \rightarrow 0} \ln [(1 + x)^{\frac{1}{x}}]} = \lim_{x \rightarrow 0} e^{\ln (1 + x)^{\frac{1}{x}}} = \lim_{x \rightarrow 0} (1 + x)^{\frac{1}{x}}[/tex]

    [tex]\boxed{e = \lim_{x \rightarrow 0} (1 + x)^{\frac{1}{x}}}[/tex]

    [tex]n = \frac{1}{x} \; \; \; n \rightarrow \infty \; \; \; x \rightarrow 0^+[/tex]

    [tex]\boxed{e = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n}[/tex]

    Is this solution correct?

    [tex]\frac{d}{dx} (\ln x) = \lim_{h \rightarrow 0} \frac{\ln(x + h) - \ln x}{h} = \lim_{h \rightarrow 0} \ln [(x + h)^{\frac{1}{h}}][/tex]

    [tex]\boxed{\frac{d}{dx} (\ln x) = \lim_{h \rightarrow 0} \ln [(x + h)^{\frac{1}{h}}]}[/tex]

    [tex]\lim_{h \rightarrow 0} \frac{1}{x} \ln(x + h) = \frac{\ln x}{x}[/tex]

    [tex]x = e[/tex]

    [tex]\frac{\ln x}{x} = \frac{\ln e}{e} = \frac{1}{e}[/tex]

    [tex]\boxed{\frac{d}{dx} (\ln x) = \lim_{h \rightarrow 0} \frac{\ln(x + h) - \ln x}{h} = \frac{1}{x}}[/tex]

     
    Last edited: Jan 21, 2006
  18. Jan 21, 2006 #17

    VietDao29

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    In all my proof, I just use the fact that xa (a is an integer) is continuous.
    Lemma 1:
    If [itex]\lim_{x \rightarrow \alpha} f(x) = L[/itex] then [itex]\lim_{x \rightarrow \alpha} f(x) ^ n = L ^ n[/itex] (n is some real number).
    Assume that f(x) > 0, for some x in the neighbourhood of [itex]\alpha[/itex].
    I will adopt the fact that fa(x) (a is an integer, positive or negative) is continuous, which can be shown by limits.
    ---------------
    Given an irrational number b. I'm going to prove that gb(x) is also continuous.
    We define a sequence of functions {un(x)}:
    [tex]u_n(x) = g ^ {j(n)} (x)[/tex], where j(n) is a function that will return a rational number in the range [itex]b \pm 10 ^ {-n}[/itex].
    So we can define [tex]g ^ b(x) = \lim_{n \rightarrow \infty} u_n(x), \ n \in \mathbb{N}[/tex].
    Since un(x) is continuous for all natural number n, gb(x) must also be continuous.
    ---------------
    From there, we can say that:
    If [itex]\lim_{x \rightarrow \alpha} f(x) = L[/itex] then [itex]\lim_{x \rightarrow \alpha} f(x) ^ n = L ^ n[/itex]. (Q.E.D)


    Lemma 2:
    (xa)' = axa - 1 (for all a).
    If a is rational then we can show that: (xa)' = axa - 1 (using lemma 1).
    If a is irrational, then define xa as above:
    [tex]x ^ a = \lim_{n \rightarrow \infty} v_n(x), \ n \in \mathbb{N}[/tex], where vn(x) can be define as:
    [tex]v_n(x) = x ^ {j(n)}[/tex], where j(n) is a function that will return a rational number in the range [itex]a \pm 10 ^ {-n}[/itex].
    [tex](v_n(x))' = (x ^ j(n))' = j(n) x ^ {j(n) - 1}[/tex].
    As j(n) will converge to a as n tends to infinity, we can say that:
    [tex](x ^ a)' = \lim_{n \rightarrow \infty} (v_n(x))' = ax ^ {a - 1}[/tex]


    Lemma 3:
    We define {on} to be:
    [tex]o_n = \left( 1 + \frac{1}{n} \right) ^ n[/tex]
    By using binomial theorem, we can say that: 2 < {on} < 3, and {on} is increasing. That means {on} will converge to some value as n tends to infinity, and we denote that value to be e.
    Proof: for the statement: 2 < {on} < 3
    [tex]o_n = 1 + n \frac{1}{n} + \frac{n(n - 1)}{2!} \frac{1}{n ^ 2} + ... + \frac{n!}{n!} \frac{1}{n ^ n}[/tex]
    [tex]1 + 1 + \frac{1}{2!} \left( 1 - \frac{1}{n} \right) + \frac{1}{3!} \left( 1 - \frac{1}{n} \right) \left( 1 - \frac{2}{n} \right) + ... + \frac{1}{n!} \left( 1 - \frac{1}{n} \right) \left( 1 - \frac{2}{n} \right) ... \left( 1 - \frac{n - 1}{n} \right)[/tex] (***)
    [tex]< 2 + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} \leq 2 + \frac{1}{2} + \frac{1}{2 ^ 2} + \frac{1}{2 ^ 3} + ... + \frac{1}{2 ^ n} \leq 3[/tex].
    From (***) we can show that {on} is increasing, and hence is lower bounded by o1 = 2.



    Lemma 4:
    [tex]\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x = e[/tex].
    Take the derivatives of that, we have (we can take derivatives of it due to lemma 2):
    [tex]\left( \left( 1 + \frac{1}{x} \right) ^ x \right)' = \left( 1 + \frac{1}{x} \right) ^ {x - 1} > 0[/tex], hence:
    [tex]\left( 1 + \frac{1}{x} \right) ^ x \right)[/tex] is increasing.
    There will exist a natural n such that n <= x <= n + 1. that means:
    [tex]\left( 1 + \frac{1}{n} \right) ^ n \right) \leq \left( 1 + \frac{1}{x} \right) ^ x \right) \leq \left( 1 + \frac{1}{n + 1} \right) ^ {n + 1} \right)[/tex]. Using Squeeze theorem we can show that:
    [tex]\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x = e[/tex] (Q.E.D)



    Lemma 5:
    [tex]e ^ x = \sum_{n = 0} ^ {\infty} \frac{x ^ n}{n!}[/tex]
    From lemma 1, we can show that:
    [tex]e ^ x = \lim_{k \rightarrow \infty} \left( 1 + \frac{x}{k} \right) ^ k[/tex]. Taking derivatives of that gives:
    [tex](e ^ x)' = \lim_{k \rightarrow \infty} \left( \left( 1 + \frac{x}{k} \right) ^ k \right)' = \lim_{k \rightarrow \infty} \left( 1 + \frac{x}{k} \right) ^ {k - 1} = e ^ x[/tex].
    Using e0 = 1, and (ex)' = ex, we can show that:
    [tex]e ^ x = \sum_{n = 0} ^ {\infty} \frac{x ^ n}{n!}[/tex] (Q.E.D).
    From here, we can show another definition for e:
    [tex]e = \sum_{n = 0} ^ {\infty} \frac{1}{n!}[/tex].



    From lemma 5, we can say that ex is continuous, and increasing. Hence it has an inverse function, denote as ln(x), whose graph is a reflection of the graph ex across the line y = x. Hence ln(x) is also continuous. Some of the proof for the properties for exp and ln are shown in my earlier post.


    Also from my earlier post here, we can say that (ln(x))' = 1 / x. From which we can show your definition for ln(x):
    [tex]\ln (x) = \int \limits_{1} ^ x \frac{dx}{x}[/tex]



    Yes, I know Vietnamese book sucks :yuck:. So please give me some advice, and opinion, so I can expand my knowledge. And maybe I may write to the publisher asking them to write clearer, more structural, and accurate books. Is there any fraud in my reasoning? (Lemma 3 is taken out of my abstract algebra book).
    Yeah, reading back my abstract algebra book, I find lots of things that are so unclear, or are claimed to be generally accepted. So I would like to ask you guys what books in English that teach us Calculus (all courses), Abstract Algebra,... with good structures, resonable and clear explanation, so that I can read them to consolidate, and expand my knowledge.
    Thanks,
     
    Last edited: Jan 21, 2006
  19. Jan 21, 2006 #18

    benorin

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    Orion1, I like what you have: very good. But I cannot figure how you got this:

     
    Last edited: Jan 21, 2006
  20. Jan 21, 2006 #19

    I can supply you with a reference for the Calculus book that my college is currently using. My college is very selective regarding their mathematics books and this is a recent publication:


    Calculus - James Stewart 5e (5th edition)
    ISBN: 0-534-39339-X

    Available for Purchase:
    http://websites.swlearning.com/cgi-...ct_isbn_issn=053439339X&discipline_number=436


    This is the circular argument:

    [tex]f(x) = \ln x[/tex]
    [tex]f'(x) = \frac{1}{x}[/tex]
    [tex]f'(1) = 1[/tex]

    The equation should actually read:
    [tex]\lim_{x \rightarrow 0} \ln [(1 + x)^{\frac{1}{x}}] = 1[/tex]
    [tex]f'(1) = 1[/tex]

    [tex]\lim_{x \rightarrow 0} \ln [(1 + x)^{\frac{1}{x}}] = \ln [\lim_{x \rightarrow 0} (1 + x)^{\frac{1}{x}}] = \ln e = 1[/tex]

     
    Last edited: Jan 21, 2006
  21. Jan 21, 2006 #20

    benorin

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    Alternatively, [tex]\exp (kx),[/tex] k fixed, is the unique continuous non-trivial solution to the functional equation [tex]f(x+y)=f(x)f(y),\forall x,y\in\mathbb{R}[/tex]
     
    Last edited: Jan 21, 2006
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