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Looking to prove $ f(x) = e^x =f'(x) $ using the limit definition of the derivative. So far I have gotten to here:

$e^x * \lim\limits_{h \to 0} \frac{e^h-1}{h}$

Any hints on what I can do next to get that entire limit to $1$?

I can't use l'hopital's rule, otherwise that'd defeat the purpose.

Ali
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Seth Killian
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    What's your definition of $e$? – Chappers May 21 '17 at 14:19
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    https://math.stackexchange.com/questions/410209/the-derivative-of-ex/1915030#1915030 –  May 21 '17 at 14:20
  • What is your definition of $\Bbb e ^x$? Many people take it to be $\sum _{n \ge 0} \dfrac {x^n} {n!}$, in which case it is quite easy to show what you want. – Alex M. May 22 '17 at 07:58

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Derivative of a function is not that difficult to calculate provided you know the definition of the function very well. Most students who start to learn calculus are aware only of the definition of polynomials, rational functions and to some extent algebraic functions. Thanks to many crappy textbooks they are led to believe that they also know the definitions of circular, logarithmic and exponential functions. Proper definitions of these functions is not possible without a proper theory of real numbers and some amount of real-analysis or calculus.

One way out (especially suitable for beginners in calculus) is not to define these complicated functions, but rather state their most common properties (including certain limits associated with them) without proof and then using these properties the derivative of these functions can be calculated. For the case of $f(x) =e^{x}$ we need to know two properties $$e^{a+b} =e^{a} e^{b}, \, \lim_{x\to 0}\frac{e^{x}-1}{x}=1$$ and using these you can easily show that the derivative of $e^{x} $ is $e^{x} $ itself.

Later when you have attained some maturity in calculus you can very well learn a proper definition of $e^{x} $ using which you can prove the properties mentioned above. Again there are many possible options to define these functions and anyone can be chosen. My own preferred definition is given by the equation $$e^{x} =\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}$$ and you can study the development of these functions based on this definition in this post.

Paramanand Singh
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    The problem with the exponential, sine and cosine functions is that you need them before the necessary tools for a rigorous definition are available. One can tell some “white lies” at the beginning, explaining later where the weak points were, when providing the “correct” definitions with power series or other methods. My preferred way is to define the logarithm with the integral of $1/x$ and the exponential as the inverse. – egreg May 21 '17 at 16:32
  • @egreg : fully agree, if you leave these functions then where are the examples of transcendentals in calculus. Perhaps you will have to be content with cheap examples like $[x] $ greatest integer function. On the other hand rarely do I find introductory calculus textbooks which are honest enough to explicitly state these properties without proof and delegate proofs to advanced courses. Most prefer to do some amount of hand waving. BTW the integral definition of $\log$ is the simplest and perhaps more analytically appealing. I also read it first and then the limit definition of $e^{x} $. – Paramanand Singh May 21 '17 at 16:39
  • @egreg : liked the use of phrase "white lies". +1 for the comment. – Paramanand Singh May 21 '17 at 16:41
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$$\large e=\lim_{\Delta x \to 0 }(1+\Delta x)^{\frac{1}{\Delta x}}\space (definition)\\ e^x=\lim_{\Delta x \to 0 }(1+\Delta x)^{\frac{x}{\Delta x}}\\ f'(x)=\lim_{\Delta x \to 0 }\frac{f(x+\Delta x)-f(x)}{\Delta x }=\\ \large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x+\Delta x}{\Delta x}}-(1+\Delta x)^{\frac{x}{\Delta x}}}{\Delta x }=\\ \large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x}{\Delta x}}((1+\Delta x)^{\frac{\Delta x}{\Delta x}}-1)}{\Delta x }=\\ \large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x}{\Delta x}}((1+\Delta x)^1-1)}{\Delta x }=\\ \large \lim_{\Delta x \to 0 }\frac{(1+\Delta x)^{\frac{x}{\Delta x}}(\Delta x)}{\Delta x }=\\ \large \lim_{\Delta x \to 0 }(1+\Delta x)^{\frac{x}{\Delta x}}=\\e^x$$

Khosrotash
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  • Is the definition for e not the limit as delta x approaches infinity rather than 0? – Seth Killian May 21 '17 at 14:29
  • $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\ (e^x)'=0+1+\frac{2x}{2!}+\frac{3x^2}{3!}+\frac{4x^3}{4!}+\frac{5x^4}{5!}+...(\text{simplify})\(e^x)'=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=e^x$$ – Khosrotash May 21 '17 at 14:32
  • @SethKillian If $\Delta x = \frac 1n$ then $\Delta x \rightarrow 0$ as $n \rightarrow \infty$ and $e = \lim_{n \rightarrow \infty}(1 + \frac 1n)^n = \lim_{\Delta x \rightarrow 0} ( 1 + \Delta x)^{\frac {1}{\Delta x}}$ – fleablood May 21 '17 at 16:27