Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
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
It may not matter which, but you need to pick one! –  Mariano Suárez-Alvarez Nov 28 '10 at 3:54
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

up vote 10 down vote accepted

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 $g(y)=1$.

share|cite|improve this answer

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.

share|cite|improve this answer
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
@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 $y=b$

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

share|cite|improve this answer

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

share|cite|improve this answer
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, $$ \lim_{a\to b} \frac{b-a}{\log b-\log a} = b $$


  • 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.
share|cite|improve this answer
(Actually I guess the argument proves FTC for any convex $f$.) –  Steven Taschuk Jun 8 '14 at 13:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.