I was trying to solve the derivative of $e^{x}$ the traditional way with the definition of the derivative: $$ \lim_{h\rightarrow 0}\frac{e^{x+h}-e^{x}}{h} $$ so I solved like this:

$$\lim_{h\rightarrow 0}\frac{e^{x+h}-e^{x}}{h} = \lim_{h\rightarrow 0}\frac{e^{x}\cdot e^{h} -e^{x}}{h}= e^{x}\cdot \lim_{h\rightarrow 0}\frac{e^{h} -1}{h}=e^{x}$$ where I solved $\lim_{h\rightarrow 0}\frac{e^{h} -1}{h}$ numerically to get 1 (which was expected because I've done this with implicit differentiation and with Taylor Series as proofs for the derivative of $e^{x}$ to be itself).

So how to solve $\lim_{h\rightarrow 0}\frac{e^{h} -1}{h}$ non-numerically and without L'Hôspital's rule?

  • 2
    $\begingroup$ I think if you are going to ask this question you need to tell us exactly how you are defining $e$. $\endgroup$
    – Seth
    Oct 17 '14 at 23:07
  • 1
    $\begingroup$ math.stackexchange.com/questions/374119/… $\endgroup$
    – nabla
    Oct 17 '14 at 23:11
  • $\begingroup$ @Seth If I don't want to run in circles I think it would be $\lim_{x\rightarrow \infty}\left ( 1+\frac{1}{x} \right )^{x}$ $\endgroup$
    – Zaid Ajaj
    Oct 17 '14 at 23:15
  • $\begingroup$ Once you take the power series expansion, you get 1 as h goes to zero. Where do you need L'Hospital's rule? $\endgroup$ Oct 17 '14 at 23:16
  • 1
    $\begingroup$ First of all, the definition is not derived using a property of what is defined. en.wikipedia.org/wiki/Exponential_function In this page 3 definitions of exponential function are given. It would be a good exercise to start from any of these definitions and derive the derivative. The easiest would be from the first definition. The equality of the three statements can be proven after getting the derivative. There is no circular argument in any of these cases. $\endgroup$ Oct 17 '14 at 23:57

Let $t = e^h -1$. Then we obtain that $h = \ln(1+t)$. Observe that $h\rightarrow 0$ implies that $t\rightarrow 0$ and we have:

$$\lim_{h\rightarrow 0} \frac{e^{x+h}-e^x}{h}= \lim_{h\rightarrow 0}\frac{e^x(e^h-1)}{h} = \lim_{t\rightarrow 0}e^x\cdot \frac{t}{\ln (1+t)}=e^x\cdot \lim_{t\rightarrow 0}\frac{1}{\frac{1}{t}\ln (1+t)}=$$ $$= e^x\cdot \frac{1}{\displaystyle \lim_{t\rightarrow 0} \ln (1+t)^{\frac{1}{t}}}= e^x\cdot \frac{1}{\ln \displaystyle \lim_{t\rightarrow 0} (1+t)^{\frac{1}{t}}}=e^x\cdot\frac{1}{\ln e} =e^x.$$

In this calculus we use the fundamental limit: $\displaystyle \lim_{t\rightarrow 0}(1+t)^{\frac{1}{t}} =e$.


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