# How to evaluate $\lim\limits_{h \to 0} \frac {3^h-1} {h}=\ln3$?

How is

$$\lim_{h \to 0} \frac {3^h-1} {h}=\ln3$$

evaluated?

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are you sure you want $h \to \infty$? I think you want $h \to 0$. Then recall de l'Hôpital - the third example given in Wikipedia carries out the details. –  t.b. May 8 '11 at 8:04
my bad. Yes it is $h\to 0$ –  amul28 May 8 '11 at 8:06
The limit is already in the form $\displaystyle{\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}}$; if you know how to find the derivative, there's no reason to use l'Hôpital. –  Jonas Meyer May 8 '11 at 8:08
@Theo Buehler: yes, i got it. So, is it something like pre-defined relation right. –  amul28 May 8 '11 at 8:11
I'm not sure I understand these last two comments. Are you asking why $f(h) = 3^{h}$ has derivative $f'(h) = \log{3}\cdot 3^{h}$? –  t.b. May 8 '11 at 8:15

There are at least two ways of doing this: Either you can use de l'Hôpital's rule, and as I pointed out in the comments the third example on Wikipedia gives the details.

I think a better way of doing this (and Jonas seems to agree, as I saw after posting) is to write $f(h) = 3^{h} = e^{\log{3}\cdot h}$ and write the limit as $$\lim_{h \to 0} \frac{f(h) - f(0)}{h}$$ and recall the definition of a derivative. What comes out is $f'(0) = \log{3}$.

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changing the function and solving seems clear for me. –  amul28 May 8 '11 at 8:24
@amul28: ok, very good, then. –  t.b. May 8 '11 at 8:28
If I remember correctly, the fact that $\lim_{x\to0}\frac{e^x-1}x=1$ is usually shown before the definition of derivative in the introductory courses. Therefore the second solution is accessible even to students that haven't heard of derivatives yet and are studying limits. –  Martin Sleziak May 8 '11 at 8:29
@Martin: Very good point. But it doesn't hurt pointing out that it comes down to taking the derivative of a very common function, does it? –  t.b. May 8 '11 at 8:32
Following up on Martin's comment, if you wanted to reduce it to the limit with $e$ instead of $3$, you could use $\displaystyle{\frac{3^h-1}{h}=\frac{e^{h\ln(3)}-1}{h}=\ln(3)\cdot\frac{e^{h\ln(‌​3)}-1}{h\ln(3)}=\ln(3)\cdot\frac{e^t-1}{t}}$, where $t=h\ln(3)$ goes to $0$ as $h$ does. –  Jonas Meyer May 8 '11 at 8:35

The result can also be obtained using $\int_a^b {e^x \,dx} = e^b - e^a$ (for all $a,b \in \mathbb{R}$). Indeed, for any $h \neq 0$ it holds $$\frac{{3^h - 1}}{h} = \frac{{\int_0^{(\ln 3)h} {e^u \,du} }}{h},$$ and hence (since $x \mapsto e^x$ is continuous and $e^0=1$) $$\mathop {\lim }\limits_{h \to 0} \frac{{3^h - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\int_0^{(\ln 3)h} {1\,du} }}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{(\ln 3)h}}{h} = \ln 3.$$

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Recall the convention $\int_a^b {f(x)\,dx} = - \int_b^a {f(x)\,dx}$. –  Shai Covo May 8 '11 at 11:37

If you want a philological answer, this limit must be proved only by means of the definition of $x \mapsto a^x$. Of course it is the definition of the derivative of this function at $x_0=0$, and therefore you should not use De l'Hospital's rule: how can you compute a derivative if you do not know how to compute the same derivative?

These limits are always troublesome, since most of us do not learn a rigorous definition of the exponential function before computing elementary derivatives. That's why the limit $$\lim_{x \to 0} \frac{e^x-1}{x}=1$$ is a fundamental limit: it is so hard to prove without a circular reasoning that calculus teachers take it for granted.

However, if you do not remember the final result, any tool is welcome, provided it gives you the correct answer!

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Am I right in my proof? –  Frank Science Jul 5 '12 at 11:37
Yes. It is also possible to note that $$\frac{a^x-1}{x}=\frac{e^{x \log a}-1}{x}$$ and multiply+divide by $\log a$. –  Siminore Jul 5 '12 at 11:51
How is philology (with phi-, not phy-) related to the question? –  Did Jul 15 '12 at 6:10

This is such a basic limit that might as well be built up before the derivative, so I think the L'Hospital or derivative had better be avoided, and the following proof is necessary:

We prove that $\lim_{x\to0}\,(a^x-1)/x=\ln a$.

First we prove that $\lim_{x\to0}\,x/\log_a(1+x)=\ln a$.

$$\lim_{x\to0}\frac{\log_a(1+x)}x=\lim_{x\to0}\log_a(1+x)^{1/x}=\log_ae$$

(Can you prove that $\lim_{x\to0}(1+x)^{1/x}=e$ and $\log$ is continuous?)

So

$$\lim_{x\to0}\frac x{\log_a(1+x)}=\ln a$$

In your problem, let $y=a^x-1$, and $x\to0$ (and never $=0$), we have $y\to0$ (and never $=0$), so $(a^x-1)/x=y/\log_a(1+y)\to\ln a$.

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Why should the limit of $(1+x)^{1/x}$ be considered as more elementary than the limit of $\log(1+x)/x$? –  Did Jul 15 '12 at 6:13
@did As far as I know, most calculus book bases the properties of $e^x$ in derivative and integration on $\lim_{x\to\infty}(1+x)^{1/x}$. More precisely, firstly we have $\lim_{n\to\infty}(1+1/n)^n$ exists, which is the definition of $e$. Secondly, we can show that $(1+1/x)^x\to e$ as $x\to+\infty$ and $x\to-\infty$. For $x>0$, we have $(1+1/r)^l\le(1+1/x)^x\le(1+l)^r$, where $l=\lfloor x\rfloor$ and $r=\lceil x\rceil$. –  Frank Science Jul 15 '12 at 10:50
@did The word should seems offensive, so I change a word. –  Frank Science Jul 15 '12 at 10:51
Sorry but this is not very convincing. The construction of real powers of positive real numbers seem to be based on, hence to be less elementary than, the log and exp functions. –  Did Jul 15 '12 at 10:55
@did But the derivative (say, $d(e^x)=e^xdx$) is no less elementary than $\log$ and $\exp$ functions. –  Frank Science Jul 15 '12 at 10:59

$\lim_{h\to 0}\frac{3^h-1}{h}=\lim_{h\to 0}\frac{e^{h\log 3}-1}{h}$. Now expansion of $e^{h\log 3}=1+\frac{h\log 3}{1!}+\frac{h^2(\log 3)^2}{2!}\cdots \implies \frac{e^{h\log 3}-1}{h}=\log 3+\frac{h(\log 3)^2}{2!}\cdots \implies \lim_{h\to 0}\frac{e^{h\log 3}-1}{h}=\log 3+0+0+\cdots = \log 3$ Hence, $\lim_{h\to 0}\frac{3^h-1}{h}=\log 3$.

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Using the formula $$\lim_{x \to 0} \frac {a^x-1} {x}=\ln a,$$ we have for $a = 3$ $$\lim_{h \to 0} \frac {3^h-1} {h}=\ln 3.$$

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-1. This precludes the desired result. –  Did Jul 16 '12 at 6:30