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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?

hjhjhj57
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Zaid Ajaj
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    I think if you are going to ask this question you need to tell us exactly how you are defining $e$. – Seth Oct 17 '14 at 23:07
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    http://math.stackexchange.com/questions/374119/can-displaystyle-lim-h-to-0-fracbh-1h-be-solved-without-circular-rea?rq=1 – nabla Oct 17 '14 at 23:11
  • @Seth If I don't want to run in circles I think it would be $\lim_{x\rightarrow \infty}\left ( 1+\frac{1}{x} \right )^{x}$ – Zaid Ajaj Oct 17 '14 at 23:15
  • Once you take the power series expansion, you get 1 as h goes to zero. Where do you need L'Hospital's rule? – Mathgrad Oct 17 '14 at 23:16
  • @MathewGeorge Doesn't that require to take derivatives? I'm assuming I don't know the derivative of $e^{x}$, that is what I'm solving for in the first place! – Zaid Ajaj Oct 17 '14 at 23:23
  • ${e^h-1}/h=1+{h/2!}+{h^2/3!}...$ which goes to 1 as h goes to 0. – Mathgrad Oct 17 '14 at 23:26
  • @MathewGeorge using the fact that $e^{x} = \sum_{n=0}^{\infty}\frac{x^{n}}{n!}$ requires knowledge of the derivative of $e^{x}$ which we are trying to solve here so we can't use that definition – Zaid Ajaj Oct 17 '14 at 23:33
  • That is how exponential function is defined. It is a definition, not a theorem based on some conditions. Other equivalent definitions can be derived after finding the derivative using this definition. – Mathgrad Oct 17 '14 at 23:42
  • @MathewGeorge It seems circular reasoning because that definition is derived from the fact that $(e^{x})'= e^{x}$ it did not come out of the blue! – Zaid Ajaj Oct 17 '14 at 23:47
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    First of all, the definition is not derived using a property of what is defined. http://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. – Mathgrad Oct 17 '14 at 23:57
  • If you define $e$ to be $\sum_{n\ge0}\frac1{n!}=\frac11+\frac11+\frac12+\frac16+\dotsb$: I think you would first need to prove that $e^x=1+x+\frac{x^2}2+\frac{x^3}6+\dotsb$. (To prove it, I think I'd define $\exp(x)=1+x+\frac{x^2}2+\frac{x^3}6+\dotsb$. Then, I'd prove that $\exp$ is continuous and convergent everywhere, that $\exp(1)=e$, and that $\exp(a)\exp(b)=\exp(a+b)$. Using those, it's not too hard to show that $\exp(x)=e^x$.) Then the limit can be evaluated easily. – Akiva Weinberger May 28 '15 at 01:05

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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: $\displaystyle \lim_{t\rightarrow 0}(1+t)^{\frac{1}{t}} =e$.

ZHN
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