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Our applied calculus text defines $e$ by $e=\displaystyle\lim_{h\to0}(1+h)^{1/h}$,

and then gives the following argument to show that $D_{x}(e^x)=e^x$:

If $f(x)=e^x$, then

$\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{e^{x+h}-e^x}{h}=\lim_{h\to0}\frac{e^x(e^h)-e^x}{h}=\lim_{h\to0}\frac{e^x(1+h)-e^x}{h}=\lim_{h\to0}\frac{e^x(h)}{h}=e^x.$

I have two questions about this argument:

1) Is there a simple way to justify replacing $e^h$ by $1+h$ in the numerator?*

2) If not, is there a way to prove that $\displaystyle \lim_{h\to0}\frac{e^h-1}{h}=0$ using the definition of $e$ given above?*

*(without using Taylor series or logarithms, neither of which have been covered at this point in the book).

user84413
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    What's your definition for $a^x$? – Zero Jan 18 '15 at 16:15
  • It might be helpful to note that $e^x=\lim_{h\rightarrow 0}(1+h)^{x/h}$ - the algebra follows pretty easily if you do this, but you end up with two nested limits. – Milo Brandt Jan 18 '15 at 16:40
  • @Zero That's a good question - the text doesn't really define $a^x$; it just assumes that the reader knows what $a^x$ means for $a>0$ and $x\in\mathbb{R}$. – user84413 Jan 18 '15 at 16:50
  • The most elementary way to define $a^x$ is as $\exp(\log(a)\cdot b)$ where $\log(x)=\int_1^x1/tdt$ and $exp=log^{-1}$ (the inverse function). So one either uses logarithms or series to be able to define $a^x$ properly. – Zero Jan 18 '15 at 16:55
  • @Zero I agree with you, but this is typical of applied calculus texts. – user84413 Jan 18 '15 at 16:57
  • what is $e$ for you? – abel Jan 18 '15 at 17:05
  • @abel: read the first line. –  Jan 18 '15 at 17:39

3 Answers3

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We are going to show that $$\lim_{h \to 0} (1+h)^\frac{1}{h}=e$$ implies $$\lim_{t \to 0} \frac{e^{t}-1}{t}=1$$

Define $h_t=e^{t}-1$. When $t \to 0$ we have $h_t \to 0$. Therefore $$\lim_{h_t \to 0} (1+h_t)^\frac{1}{h_t}=e$$ $$\lim_{t \to 0} (e^{t})^\frac{1}{e^{t}-1}=e$$

$$\lim_{t \to 0} e^\frac{t}{e^{t}-1}=e$$

Note that in general $\lim_{t \to a} f(g(t))=f(c)$ doesn't imply that $\lim_{t \to a} g(t)=c$. BUT, in this case the claim can be proven using the fact that the exponential is strictly increasing and continuous at $0$.

N. S.
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Let us define the exponential using integer powers by $$e^x=\lim_{n\to\infty}(1+\frac xn)^n.$$ This definition is consistent with that of $e$, $$e=\lim_{n\to\infty}(1+\frac1n)^n=\lim_{h\to0}(1+h)^{1/h}.$$ Then, using the ordinary binomial theorem $$(1+\frac xn)^n =\sum_{k=0}^n\binom nk\left(\frac xn\right)^k =1+\frac nnx+\frac{n(n-1)}{2n^2}x^2+\frac{n(n-1)(n-2)}{2.3.n^3}x^3+\cdots+\frac{x^n}{n^n}.$$ All coefficients are positive and bounded by $1$, hence $$1+x\le\left(1+\frac xn\right)^n\le\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}\le\frac1{1-x},$$ for $0\le x<1$.

In the limit $n\to\infty$, $$1+x\le e^x\le\frac1{1-x}.$$ Then $$1\le\frac{e^x-1}x\le\frac1{1-x},$$

and the limit for $x\to0$ is clearly $1$. Similar bracketing holds for $x<0$.

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Note you can view $\displaystyle \lim_{h\to 0} \dfrac{e^h-1}{h} = (e^h)'(0)=1$

DeepSea
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