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In Calculus, the Equation is known as:

$$f'(x)=\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

This equation allow us to find the derivatives of functions. Let's try this with the exponential function:$f(x)=a^x$ where $a \gt 1$.

$$f'(x)=\lim\limits_{h \to 0} \frac{a^{x+h}-a^x}{h}$$ $$f'(x)=\lim\limits_{h \to 0} \frac{a^x a^h-a^x}{h}$$ $$f'(x)= a^x \lim\limits_{h \to 0} \frac{a^h-1}{h}$$

As you can see, we need to determine $\lim\limits_{h \to 0} (\frac{a^h-1}{h})$ to find the derivative. At this point, we ask the question: what would $a$ be such that the limit would be 1? The answer is $e$ and that's the end of everything.

What I want to know is can we actually evaluate that limit? How can we actually find $e$ other than guessing with trial and error?

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You can use binom theory to find taylor expension result of $$ \lim\limits_{h \to 0} \frac{a^h-1}{h}$$

if $a=k+1$

$(1+k)^h=1+C(h,1)k+C(h,2)k^2+C(h,3)k^3+.....$

$(1+k)^h=1+hk+\frac{h(h-1)}{2!}k^2+\frac{h(h-1)(h-2)}{3!}k^3+.....$

$$g(k)= \lim\limits_{h \to 0} \frac{(1+k)^{h}-1}{h}=\frac{hk+\frac{h(h-1)}{2!}k^2+\frac{h(h-1)(h-2)}{3!}k^3+.....}{h}=\lim\limits_{h \to 0} (k-\frac{k^2}{2}+\frac{k^3}{3}-\frac{k^4}{4}+....)+hU_1(k)+h^2U_2(k)+....$$

$$ \lim\limits_{h \to 0} \frac{(1+k)^{h}-1}{h}=k-\frac{k^2}{2}+\frac{k^3}{3}-\frac{k^4}{4}+....$$ $$ g(k)=k-\frac{k^2}{2}+\frac{k^3}{3}-\frac{k^4}{4}+....$$

$$ g(a-1)=\lim\limits_{h \to 0} \frac{a^h-1}{h}$$

$$ g(b-1)=\lim\limits_{h \to 0} \frac{b^h-1}{h}$$

$$ g(ab-1)=\lim\limits_{h \to 0} \frac{(ab)^h-1}{h}=\lim\limits_{h \to 0} \frac{(ab)^h-a^h+a^h-1}{h}=\lim\limits_{h \to 0} \frac{a^h(b^h-1)}{h}+\lim\limits_{h \to 0} \frac{a^h-1}{h}=\lim\limits_{h \to 0} \frac{(b^h-1)}{h}+\lim\limits_{h \to 0} \frac{a^h-1}{h}=g(b-1)+g(a-1)$$

We have a relation for $g(x)$ function. $$ g(ab-1)=g(b-1)+g(a-1)$$

and if you put $b=1/a$

$$ g(a^{-1}-1)=-g(a-1)$$

It is easy to show $g(0)=0$

you want to find $g(e)=1$ thus

First try to find $g(1)$ ($ g(1)= \lim\limits_{h \to 0} \frac{2^h-1}{h}$

$$ g(1)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....$$ even you calculate first 4 terms it will be near to $0.58$ (it is about 0.69)

$g(1)\approx 0.69$

$$ g(2.2-1)=g(1)+g(1)$$ $$ g(3)=g(1)+g(1)\approx 1.38$$ thus we can see that $2<e<4$

$$g(1/2)=\frac{1}{2}-\frac{1}{4.2}+\frac{1}{8.3}-\frac{1}{16.4}+....$$

$$g(1/2)\approx 0.4$$

To use $$ g(ab-1)=g(b-1)+g(a-1)$$ again $$ g(3.3/2.2 -1)=g(3/2-1)+g(3/2-1)$$ $$ g(9/4 -1)=g(3/2-1)+g(3/2-1)$$ $$ g(9/4 -1)=2g(1/2)$$

$$ g(5/4 )=2g(1/2) \approx 0.8$$

It means $e>5/4+1=2.25$

You can try more values to get near to e value.

You can use with such Technics to find e value range

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  • $\begingroup$ Thank you, this is exactly what I'm looking for! $\endgroup$ – ChaoSXDemon Aug 24 '15 at 16:34
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You got the expression

$$a^x \lim\limits_{h \to 0} \frac{a^h-1}{h}.$$

We make the change of variables $h=\log_a x$ (see that $a>1$). Then $$ \lim\limits_{h \to 0} \frac{a^h-1}{h}=\log a\lim_{x\to 1}\frac{x-1}{\log x}=\log a\frac{1}{\lim_{x\to 1}\frac{\log x}{x-1}}. $$ Now, this limit is equal to $$ \lim_{x\to 1}\frac{\log x}{x-1}=\lim_{x\to 1}\log(x^{1/(x-1)}). $$ Doing the change $x=t+1$ we arrive to $$ \lim_{x\to 1}\frac{\log x}{x-1}=\lim_{t\to 0}\log(t+1)^{1/t}=1. $$

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  • $\begingroup$ I thought only the natural log would have the above expression evaluate to 1 and therefore, $f(x)=e^x$ results in $f'(x) = e^x$. What happen to the base $a$? Are we allow to simply carry operators (taking logs) into limit operators? $\endgroup$ – ChaoSXDemon Aug 24 '15 at 5:37
  • $\begingroup$ The written log's without basis are natural log's. I change the basis a using the property $ \log_a x=(\log x)/(\log a) $ (again, natural log's) $\endgroup$ – MrSelberg Aug 24 '15 at 5:42
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One of the several definitions of $e$ is: $$e = \lim_{x \to 0} (1 + x)^{1/x}.$$ Taking the natural log of both sides and interchanging the log and the limit (which is allowed by definition of continuity), $$1 = \ln\left(\lim_{x \to 0} (1 + x)^{1/x}\right) = \lim_{x \to 0}\left( \ln (1 + x)^{1/x}\right) = \lim_{x \to 0} \frac{\ln(1 + x)}{x} = \lim_{x \to 0} \frac{x}{\ln(1 + x)}$$

Let $y = a^h - 1$. Then $h = \frac{\ln(y+1)}{\ln a}$, so $$\lim_{h \to 0} \frac{a^h - 1}{h} = \lim_{y \to 0} \frac{y \ln a}{\ln(y+1)} = \ln a \cdot \lim_{y \to 0} \frac{y}{\ln(y+1)} = \ln a.$$

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  • $\begingroup$ While this is a great answer, why would someone without knowing the magic of $e$ suddenly to and use the definition of $e$? Furthermore, I feel like this is a guided answer in the sense that we know we want to find $lna$. Am I being confusing? $\endgroup$ – ChaoSXDemon Aug 24 '15 at 5:35
  • $\begingroup$ It's all based on definitions. Some books define $e$ as the unique number such that $$\lim_{h \to 0} \frac{e^h - 1}{h} = 1.$$ Traditionally, the natural log isn't even defined to be the inverse of the exponential function. The classical approach is: $$\ln(x) = \int_1^x \frac{\mathrm{d}t}{t}.$$ Then one would prove that they are indeed inverses. My suggestion would be to not fret over these details. Just accept whichever definitions you want to accept, maybe spend some time proving their equivalence, and move on. $\endgroup$ – user217285 Aug 24 '15 at 7:49
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Notice, $$a^h=1+\frac{h}{1!}(\log a)+\frac{h^2}{2!}(\log a)^2+\frac{h^3}{3!}(\log a)^3+\ldots$$ Now, we have $$\lim_{h\to 0}\frac{a^h-1}{h}$$ $$=\lim_{h\to 0}\frac{\left(1+\frac{h}{1!}(\log a)+\frac{h^2}{2!}(\log a)^2+\frac{h^3}{3!}(\log a)^3+\ldots\right)-1}{h}$$ $$=\lim_{h\to 0}\frac{\left(\frac{h}{1!}(\log a)+\frac{h^2}{2!}(\log a)^2+\frac{h^3}{3!}(\log a)^3+\ldots\right)}{h}$$ $$=\lim_{h\to 0}\left(\frac{1}{1!}(\log a)+\frac{h}{2!}(\log a)^2+\frac{h^2}{3!}(\log a)^3+\ldots\right)$$ $$=\lim_{h\to 0}\left(\frac{1}{1!}(\log a)+0\right)=\log a$$

Now, the limit will be $1$ if we have $$\log a=1$$$$\iff a=e^1=e$$

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    $\begingroup$ This answer is not acceptable as you have used the Taylor Series to expand $a^h$. This requires you to know the derivatives of $a^h$ which is something we are trying to find! $\endgroup$ – ChaoSXDemon Aug 24 '15 at 4:50
  • $\begingroup$ Alright, but you did not mention this in the question, can we apply L-Hospital's rule? $\endgroup$ – Harish Chandra Rajpoot Aug 24 '15 at 4:52
  • $\begingroup$ My bad for not mentioning it. Wouldn't using L-Hospital's rule also requires $f'(x)$? If so then no. If somehow you can use L-Hospital's rule without actually require to find the derivative of $a^h$ (keep it as symbolically the derivative) then I suppose ... it's okay? $\endgroup$ – ChaoSXDemon Aug 24 '15 at 4:53
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From $f'(x)= a^x \lim\limits_{h \to 0} \frac{a^h-1}{h} $, $f'(x) =a^x f'(0) $, since $f(0) = 1$, so $\lim\limits_{h \to 0} \frac{a^h-1}{h} =\lim\limits_{h \to 0} \frac{f(h)-f(0)}{h} =f'(0) $.

The "natural" solution is the one where $f'(0) = 1$, and the other answers show a variety of ways to conclude that the value that does this is $a=e =\lim_{n \to \infty} (1+\frac1{n})^n =\lim_{x \to 0} (1+x)^{1/x} $.

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Yet another way is to evaluate the limit of interest is to invoke the definition for $e^x$ expressed as

$$e^x=\lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n \tag 1$$

Using $(1)$, we can write

$$\lim_{h\to 0}\frac{a^h-1}{h}=\lim_{h\to 0}\lim_{n\to \infty}\frac{\left(1+\frac{h\log a}{n}\right)^n-1}{h} \tag 2$$

Next, we recall from the Mean Value Theorem that there exists a number $\xi$, ($0<\xi<h\log a/n$ for $a>1$, $h\log a/n<\xi<0$, for $0<a<1$),such that

$$\left(1+\frac{h\log a}{n}\right)^n=1+h\,\log a+\frac{n(n-1)}{2n^2}(1+\xi)^{n-2} (h\log a)^2 \tag 3$$

Therefore, using $(3)$ in $(2)$ reveals

$$\begin{align} \lim_{h\to 0}\lim_{n\to \infty}\frac{\left(1+\frac{h\log a}{n}\right)^n-1}{h}&=\log a+\lim_{h\to 0}\lim_{n\to \infty}\frac{n(n-1)}{2n^2}(1+\xi)^{n-2} h\log^2 a\\\\ &=\log a \end{align}$$

as was to be shown!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Aug 28 '15 at 19:22

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