Derivation for the derivative of $a^{t}$ from The Equation

Question

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?

Answer 1

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

Answer 2

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 $$\log a=1$$$$\iff a=e^1=e$$

Answer 3

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 $$ \lim_{x\to 1}\frac{\log x}{x-1}=\lim_{t\to 0}\log(t+1)^{1/t}=1. $$

Answer 4

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

Answer 5

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} $.

Answer 6

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!