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Where can I find info on: limit as h→0 of (a^h - 1)/h = 1

  1. Mar 31, 2013 #1
    [tex]\lim_{h\rightarrow 0 } \frac{a^{h}-1}{h} = 1[/tex]

    I'm trying to understand a proof for derivatives of exponential functions, but I'm getting stuck when I get to this property within the proof steps. What is the name of this property or where can I find a breakdown of it?
     
  2. jcsd
  3. Mar 31, 2013 #2

    Stephen Tashi

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    For an arbitrary number [itex] a [/itex] it is not true that [tex]\lim_{h\rightarrow 0 } \frac{a^{h}-1}{h} = 1[/tex]. That limit is 1 when [itex] a = e [/itex]. Are you asking for a proof of this? I think it amounts to a definition, not a theorem. One definition of [itex] e [/itex] is that it is the value of [itex] a [/itex] that makes the limit = 1.

    Of course, one can make useless definitions and there are issues that need to be proven. One must show that the limit exists in order to show that [itex] e [/itex] exists.
     
  4. Mar 31, 2013 #3
    Okay, is this then solvable with calculus 1 knowledge? Like dividing the numerator and denominator by a certain number or using L'hopital's rule?

    Since not 1, is there an answer to this?
     
  5. Mar 31, 2013 #4
    Suppose I want to find the derivative of 5^x, then

    [itex]\lim_{h\rightarrow 0 }\frac{f(x+h)-f(x)}{h} \Rightarrow
    lim_{h\rightarrow 0 }\frac{5^{x+h}-5^{x}}{h}\Rightarrow
    lim_{h\rightarrow 0 }\frac{5^{x}5^{h}-5^{x}}{h}\Rightarrow
    (5^{x})lim_{h\rightarrow 0 }\frac{5^{h}-1}{h}[/itex]

    Then what can be done from there to arrive at the derivative of 5^x?

    If I apply L'hopital, I run into another exponential function, 5^h, and it almost looks like it will turn out to be a never ending L'hopital attempt.
     
  6. Mar 31, 2013 #5

    jbunniii

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    For ##a > 0##,
    $$a^h = \left(e^{\log a}\right)^h = e^{h \log a}$$
    So by L'Hopital's rule,
    $$\begin{align}
    \lim_{h \rightarrow 0} \frac{a^h - 1}{h} &=
    \lim_{h \rightarrow 0} \frac{ e^{h \log a} - 1}{h} \\
    \\
    &= \lim_{h \rightarrow 0} \frac{(\log a) e^{h \log a}}{1} \\
    \\
    &= \log a
    \end{align}$$
     
  7. Mar 31, 2013 #6
    Jbunniii,

    When you write log a, should I understand it to be "log(base e) of a," as in lna or "natural log a?" The conversion you've written looks familiar. If so, please let me know so I can review some of the log rules on my own and come back to this overall limit problem as soon as possible, thank you.

    P.S. Please forgive me if there is an obvious answer to this specific question on your feedback, today's my first day being able to look at some math this year. Unfortunately, having to work too many hours at my non math related field of employment this year.
     
  8. Mar 31, 2013 #7

    jbunniii

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    Yes, that's right. I meant log base ##e##, the natural log.
     
  9. Apr 1, 2013 #8
    Can you confirm that this way of getting to lna is correct? Did I arrive at your same answer by coincidence or did I correctly apply the calculus rules (chain rule, product rule etc)?

    [tex]\lim_{h\rightarrow 0 } \frac{a^{h}-1}{h} = \lim_{h\rightarrow 0 } \frac{e^{h log_e a}-1}{h} = \lim_{h\rightarrow 0 } \frac{\frac{d}{dh}e^{h log_e a}-\frac{d}{dh}1}{h'} = \lim_{h\rightarrow 0 } \frac{\frac{d}{dh}e^{h log_e a}-0}{1} [/tex]

    then chain rule on
    [tex]\frac{d}{dh}e^{h log_e a} = e^{h log_e a}(hlog_e a)'[/tex]

    then product rule within bracket

    [tex](e^{hlog_e a})[(h)'(log_e a)+(h)(log_e a)'] [/tex]

    [tex]= e^{hlog_e a}[(1)lna+h(\frac{1}{a})(a)'] =(e^{hlog_e a})[lna+(h\frac{1}{a}(0))] = (e^{hlog_e a})lna[/tex]

    so then
    [tex] \lim_{h\rightarrow 0 } (e^{hlog_e a})lna = (e^{(0)log_e a})(lna) = (e^{0})lna = (1)lna = lna[/tex]

    Also, is this the same method of arriving at lna as you expressed in post #5, or are there other "sub-steps" you would have used (besides chain rule etc) instead?

    Sincerely
     
    Last edited: Apr 1, 2013
  10. Apr 1, 2013 #9

    jbunniii

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    Yes, everything you did is fine. You can save a few steps by avoiding the product rule:
    $$\frac{d}{dh}(hlog_e a) = \log_e a$$
    because ##\log_e a## is just a constant when we differentiate with respect to ##h##. Similarly, you wouldn't use the product rule to find ##\frac{d}{dh} (2h)## - the answer is clearly ##2##. However, if you use the product rule, you will still get the right answer. It just takes more steps.
     
  11. Apr 1, 2013 #10

    Mark44

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    Pretty much a nit, but ##\Rightarrow## ("implies") is not the symbol to use above. The appropriate symbol is "=", since these are all equal expressions. The implication arrow is used between statements such as equations or inequalities.
     
  12. Apr 3, 2013 #11
    Thanks for the note on using correct notation Mark44. I'll implement your advice going forward.

    For this particular proof, (btw, I haven't yet learned out to make spaces between words in the latex language):


    Definition: If [itex]f'(x)=a^{x}[/itex], then [itex]f'(0)a^{x}[/itex] (*per edit - spacing fixed)


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

    [tex]= \lim_{h\rightarrow 0 } \frac{a^{x}a^{h}-a^{x}}{h} = a^{x} * \lim_{h\rightarrow 0 } \frac{a^{h}-1}{h}[/tex]

    [tex]= a^{x} * f'(0). [/tex]


    The part that seems to make too large a leap in logic for me is how it is decided that:

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

    I just don't know what [*this specific part of] the proof is conveying, or how those two expressions are even equal. How does f'(0) equal that limit. How is that conclusion made? F prime of what? x? What might I not be seeing, or what side-notes, properties, or in-between steps could help make this equality (and proof) make more sense.

    P.S. What would be the way to make spaces between words in latex language? This way I can come back and edit this post with the correct notation.

    Sincerely
     
    Last edited: Apr 3, 2013
  13. Apr 3, 2013 #12

    Stephen Tashi

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    Instead of using the tags "tex" and "/tex" in brackets, use the tags "itex" and "/itex" when you want the latex to appear within a line of text. You can put spaces before the first tag and after the last tag in a formula.

    When you need a space within a mathematical formula inside latex tags, use "\ " (backslash and then a space).
     
  14. Apr 3, 2013 #13

    jbunniii

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    Is there a typo here somewhere? I'm guessing you meant to say, "if ##f(x) = a^x##, then ##f'(x) = f'(0) a^x##.

    Assuming that is the case:
    Just use the definition of the derivative of a function ##f## at an arbitrary point ##x##:
    $$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$$
    Now evaluate this at ##x = 0##:
    $$f'(0) = \lim_{h \rightarrow 0} \frac{f(h) - f(0)}{h}$$
    For the special case ##f(x) = a^x##, we have ##f(0) = a^0 = 1##, so
    $$f'(0) = \lim_{h \rightarrow 0} \frac{a^h - 1}{h}$$
     
  15. Apr 3, 2013 #14

    dextercioby

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    Define e and ex by the power series. It's trivial then to prove the limit for a=e then for general a.
     
  16. Apr 3, 2013 #15
    dextercioby, I haven't yet learned about power series. Going to be in calc 2 in the fall semester (Aug-Jan 2013). But hope to learn about power series before then. Thanks.
    __________________________________

    Thanks Stephen Tashi, now able to separate words while using the latex language. Also, I noticed that itex and tex dispay the limit notation differently as well.

    Using itex: [itex]\lim_{h\rightarrow 0 } \frac{f(x+h)-f(x)}{h}[/itex]

    Using tex: [tex]\lim_{h\rightarrow 0 } \frac{f(x+h)-f(x)}{h}[/tex]

    I prefer how the limit is displayed with tex, but tex is definately limiting (no pun intended) when communicating words with math symbols, so guess I'll learn over time when to use one and when to use the other.
    ___________________________________

    Thanks jbunniii, post #13 cleared that right up. I watched a video that seemed to point out that the significance of expressing the definition with f'(0) is simply to show that the derivative of [itex]a^{x}[/itex], which is [itex]a^{x}(f'(0))[/itex] always contains a multiple of itself. Would there be any other reason besides pointing out that fact for proving and defining the derivative of the exponential function this way: [itex]\frac{d}{dx}a^{x}=a^{x}(f'(0))[/itex] rather than this way: [itex]\frac{d}{dx}a^{x}=a^{x}lna[/itex]? The latter way seems more straight forward, so what other point might there be for substituting and defining with f'(0)?

    P.S. thanks for noticing the mistake I made of leaving out the "f'(0)=" definition.
     
    Last edited: Apr 3, 2013
  17. Apr 3, 2013 #16
    If I were to seek instruction, info, or resources on how to convert from

    [tex]\lim_{n\rightarrow ∞ } (1+\frac{1}{n})^{n} = \lim_{h\rightarrow 0 } (1+h)^{\frac{1}{h}} [/tex]

    could I ask within this thread since it's sort of under the umbrella topic of formal definition of derivatives for exponential functions (this being the special case e), or would it be best practice to begin a new thread?
     
    Last edited: Apr 3, 2013
  18. Apr 4, 2013 #17

    Stephen Tashi

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    I'd say it's best to explain what generality covers this situation and then start a new thread about the generality. The genearlity is how take the limit of a composition of functions. (Some would say "change of variables", but that is a catch-all term with many different meanings.)

    A tame version of the generality is (I think):

    If [itex] g(y) [/itex] s a continuousI function at the point [itex] y = b [/itex] and [itex] f(x) [/itex] is a continuous function at the point [itex] x = A. [/itex] and [itex] \lim_{y \rightarrow b}g(y) = A [/itex] then
    [itex] lim_{y \rightarrow b } f(g(x)) = \lim_{ x \rightarrow A} f(x) [/itex]

    This version is tame since I didn't say anything about [itex] \infty [/itex]. Books differ in how they treat limits that involve "[itex] \rightarrow \infty [/itex] and limiting values that "are infinite". Some try to cover these cases by the convention that a letter like "b" or "'A" can represent "infinity". Some state the cases involving infinities as separate theorems. Also when some books say two limits "are equal" they mean that the limits are the same number when both exist and both fail to exist when one fails to exist.

    In my statement, I mean all the letters to be finite quantities, so it doesn't quite apply to your question, but at least it conveys the general pattern. Your question needs a version with:

    [itex] g(h) = 1/h [/itex]
    [itex] b = \infty [/itex]
    [itex] A = 0 [/itex]
    [itex] f(n) = (1+\frac{1}{n})^{n} [/itex]

    and the desired conclusion is:

    [itex] \lim_{h \rightarrow b} f(g(h)) = \lim_{h \rightarrow \infty} (1 + \frac{1}{{1/h}})^{\frac{1}{h} }[/itex]
    [itex] =\lim_{h \rightarrow \infty} (1 + \frac{1}{h})^{\frac{1}{h}} [/itex]
    [itex] = \lim_{n \rightarrow A} f(n) = \lim_{n \rightarrow 0} (1 + \frac{1}{n})^n [/itex]
     
  19. Apr 4, 2013 #18

    jbunniii

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    That's a good question. I can't say for sure what point the video was making, but note that the proof that ##\frac{d}{dx}a^x = a^x f'(0)## did not explicitly require us to know anything about logarithms, nor indeed very much about exponential functions. The only facts that it made use of are that ##f## is differentiable and that it satisfies the following properties: ##f(x+h) = f(x)f(h)## for all ##x## and ##h##, and ##f(0) = 1##. If these conditions are satisfied, then ##f'(x) = f(x)f'(0)##. If we further assume that ##f'(0) = 1##, then ##f## it its own derivative: ##f'(x) = f(x)##.

    Indeed, we may use that last fact as a starting point to define the exponential function. Let us assume that ##f## is a differentiable function satisfying ##f'(x) = f(x)## and ##f(0) = 1##. Then by the product rule,
    $$\frac{d}{dx}(f(x)f(-x)) = f'(x)f(-x) - f(x)f'(-x) = f(x)f(-x) - f(x)f(-x) = 0$$
    so ##f(x)f(-x)## equals some constant for all ##x##. Plugging in ##x = 0##, we see that this constant must be ##f(0)f(-0) = f(0)f(0) = 1##. So ##f(x)f(-x) = 1## for all ##x##. This shows that ##f(x)## is never zero for any ##x##, and also that ##f(-x) = 1/f(x)##.

    Now suppose we have a second function ##g## satisfying the same conditions: ##g'(x) = g(x)## and ##g(0) = 1##. Then ##g## is also never zero, so we may form the quotient ##f(x)/g(x)##, and we may differentiate it using the quotient rule:
    $$\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)} = \frac{g(x)f(x) - f(x)g(x)}{g^2(x)} = 0$$
    Therefore ##f(x)/g(x)## is constant, and once again plugging in ##x = 0##, we find that the constant must equal ##f(0)/g(0) = 1/1 = 1##, which means ##f(x) = g(x)## for all ##x##. This shows that if there is a differentiable function satisfying ##f'(x) = f(x)## and ##f(0) = 1##, then it is unique. We may similarly show that ##f(x+y) = f(x)f(y)## for all ##x## and ##y##.

    So in this way, we see that all of the key properties of the exponential function follow from the simple criteria ##f'(x) = f(x)## and ##f(0) = 1##.

    On a technical note, observe that we have only shown that if a function exists which satisfies the criteria, then it is unique and has the indicated properties. There is still some work to be done to show that such a function exists if we want to have a rigorous definition of the exponential function. This can be done many ways: for example, one can define a particular power series and show that it satisfies the criteria:
    $$e^x = \sum_{n = 0}^{\infty} \frac{x^n}{n!}$$
     
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