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How can we find $\displaystyle \lim_{y\to{b}}\frac{y-b}{\ln{y}-\ln{b}}$ without using:

(a) L'Hôpital's rule, (b) the limit $\displaystyle \lim_{h \to 0}\frac{e^h-1}{h} = 1$, and (c) the fact that $\displaystyle \frac{d}{dx}\left(e^x\right) = e^x$.

The reason for the conditions is that with this limit I'm trying to prove (c), and I've done so with (b) and I gather it would be circular to use (a). So that's that. Also, I would appreciate if you could share one or more ways of proving that the derivative of $e^x$ is $e^x$. Thanks a lot for your time.

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4  
If you want to compute the derivative of $e^x$, you need to tell us what definition of $e^x$ you are using. Likewise, to compute the limit you want, you need to tell us what definition of $\ln$ you are using. –  Mariano Suárez-Alvarez Nov 28 '10 at 3:46
    
Well, any would do. But I was cautious of using $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$. Should I be? –  De Moivre Nov 28 '10 at 3:53
1  
It may not matter which, but you need to pick one! –  Mariano Suárez-Alvarez Nov 28 '10 at 3:54
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Ok. Let's pick $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$ –  De Moivre Nov 28 '10 at 3:58
    
For the limit, let's pick $\displaystyle \ln{x} = \int_{1}^{x}\frac{1}{t}\;{dt}$. –  De Moivre Nov 28 '10 at 4:06

5 Answers 5

Let us look at the limit in the title and the definition of $\log x$ (please excuse me for using $\log$ notation instead of $\ln$) as in your comment, $$\log x =\int_1^x \frac{1}{t}dt$$ By the fundamental theorem of calculus we have $$\frac{d}{dx}\log x = \frac{1}{x}$$ on the other hand $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$$ hence $$\lim_{y\to x}\frac{y-x}{\log y-\log x}=\lim_{y\to x}\frac{1}{\frac{\log y-\log x}{y-x}}=\frac{1}{1/x}=x$$ where in the last step we used the quotient rule $$\lim_{y\to a}\,g(y)=A\, \text{ and }\,\lim_{y\to a}\,h(y)=B\ne0\text{ implies }\lim_{y\to a}\,\frac{g(y)}{h(y)}=\frac{A}{B}$$ with $g(y)=1$.

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If $e^x=\sum_{n\geq0}x^n/n!$, then you can show easily that $e^{x+y}=e^xe^y$. Then $$\frac{e^{x+h}-e^x}{h} = \frac{e^h-1}{h}e^x,$$ and to conclude that $\frac{d}{dx}e^x=e^x$ we need only then show that $$\frac{e^h-1}{h}\to1$$ if $h\to0$. Using your definition, that is easy.

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Thank you. My only question is this: since $e^x = \sum_{n\ge{0}}\frac{x^n}{n!}$ is derived using $\frac{d}{dx}{e^x} = e^x$ at the expansion of Taylor's series around 0, is using it to prove that $\frac{d}{dx}{e^x} = e^x$ by any means circular? –  De Moivre Nov 28 '10 at 4:13
    
That depends on how you define $e^x$. –  Yuval Filmus Nov 28 '10 at 5:04
    
I think you can extract Mariano's definition form $e^x = \lim_{n\rightarrow\infty} (1+x/n)^n$. –  Yuval Filmus Nov 28 '10 at 5:05
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@De Moivre, @Yuval: I am defining the exponential using the series.n De Moivre, I asked you to pick a definition, and you picked that one! –  Mariano Suárez-Alvarez Nov 28 '10 at 6:07
    
@Mariano: I meant to say that even if you define $e^x$ as the limit I wrote, then you could conclude your definition. So the proof works for two definitions at once! –  Yuval Filmus Nov 28 '10 at 6:36

We can convert your $\lim $ into the inverse of the derivative of $f(y)=\ln y $, evaluated at $y=b$

$$\underset{y\rightarrow b}{\lim }\dfrac{y-b}{\ln y-\ln b}=\dfrac{1}{\underset{ y\rightarrow b}{\lim }\dfrac{\ln y-\ln b}{y-b}}=\dfrac{1}{\dfrac{d}{dy}\left. \ln y\right\vert _{y=b}}=\dfrac{1}{\left. \dfrac{1}{y}\right\vert _{y=b}}=% \dfrac{1}{\dfrac{1}{b}}=b.$$

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Say $y = b(1+\epsilon)$. Then $\ln y - \ln b \approx \epsilon$ whereas $y - b = b\epsilon$. So it all boils down to showing $\ln (1+\epsilon) \approx \epsilon$.

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Well, you need to show that $\ln y-\ln b\approx \epsilon$... –  Mariano Suárez-Alvarez Nov 28 '10 at 3:50

If $f$ is convex on $[a,b]$ or $[b,a]$ then $$ f\Big(\frac{a+b}2\Big) \le \frac1{b-a}\int_a^b f(x)\,dx \le \frac{f(a)+f(b)}2 \tag{$\ast$} $$ (The first inequality is by Jensen, since $\frac{a+b}2=\frac1{b-a}\int_a^b x\,dx$; the second comes from the change of variables $x=(1-t)a+tb=a+t(b-a)$ and applying the convexity of $f$ pointwise.)

Taking $f(x)=\frac1x$ and taking reciprocals throughout yields $$ \frac1{\frac12(\frac1a+\frac1b)} \le \frac{b-a}{\log b-\log a} \le \frac{a+b}2 $$ which is the inequality of the harmonic, logarithmic, and arithmetic means. By squeezing, $$ \lim_{a\to b} \frac{b-a}{\log b-\log a} = b $$

Notes:

  • In fact the logarithmic mean is also bounded by the geometric mean, i.e., $$ \sqrt{ab} \le \frac{b-a}{\log b - \log a} $$ This is stronger than the bound by the harmonic mean proved above, but the only proof I know is to take $f(x)=e^x$ in ($\ast$), and to evaluate the resulting integral we need to use $\frac{d}{dx} e^x = e^x$.
  • An advantage (?) of this proof is that it doesn't need the fundamental theorem of calculus. (In fact it verifies FTC for $\int\frac1x\,dx$.) That's assuming you define $\log$ as the integral of $\frac1x$, that you prove a change of variables theorem for linear changes of variables directly by Riemann sums, and that you evaluate $\int_a^b x\,dx$ directly by Riemann sums.
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(Actually I guess the argument proves FTC for any convex $f$.) –  Steven Taschuk Jun 8 at 13:53

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