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How is

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

evaluated?

  • 1
    are you sure you want $h \to \infty$? I think you want $h \to 0$. Then recall [de l'Hôpital](http://en.wikipedia.org/wiki/L'H%F4pital's_rule) - the third example given in Wikipedia carries out the details.2011-05-08
  • 0
    my bad. Yes it is $h\to 0$2011-05-08
  • 8
    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.2011-05-08
  • 0
    @Theo Buehler: yes, i got it. So, is it something like pre-defined relation right.2011-05-08
  • 0
    @Jonas Meyer: i did not get how to solve in that manner and so finding for an alternative.2011-05-08
  • 1
    I'm not sure I understand these last two comments. Are you asking why $f(h) = 3^{h}$ has derivative $f'(h) = \log{3}\cdot 3^{h}$?2011-05-08
  • 0
    no, i know that derivative of $3^h=3^h.ln3$. But i'm trying to figure it out how to slove it by $1^{st}$ order derivation.2011-05-08

6 Answers 6

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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 $f'(0) = \log{3}$.

  • 0
    changing the function and solving seems clear for me.2011-05-08
  • 0
    @amul28: ok, very good, then.2011-05-08
  • 5
    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.2011-05-08
  • 2
    @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?2011-05-08
  • 5
    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.2011-05-08
5

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 $e^0=1$) $$ \mathop {\lim }\limits_{h \to 0} \frac{{3^h - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\int_0^{(\ln 3)h} {1\,du} }}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{(\ln 3)h}}{h} = \ln 3. $$

  • 0
    Recall the convention $\int_a^b {f(x)\,dx} = - \int_b^a {f(x)\,dx} $.2011-05-08
3

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!

  • 0
    Am I right in my proof?2012-07-05
  • 0
    Yes. It is also possible to note that $$\frac{a^x-1}{x}=\frac{e^{x \log a}-1}{x}$$ and multiply+divide by $\log a$.2012-07-05
  • 2
    How is *philology* (with phi-, not phy-) related to the question?2012-07-15
2

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 $(a^x-1)/x=y/\log_a(1+y)\to\ln a$.

  • 3
    Why should the limit of $(1+x)^{1/x}$ be considered as more elementary than the limit of $\log(1+x)/x$?2012-07-15
  • 0
    @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 $r=\lceil x\rceil$.2012-07-15
  • 0
    @did The word *should* seems offensive, so I change a word.2012-07-15
  • 0
    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.2012-07-15
  • 0
    @did But the derivative (say, $d(e^x)=e^xdx$) is no *less* elementary than $\log$ and $\exp$ functions.2012-07-15
  • 0
    ?? (Anyway, I think I stated the remark I wanted to state, so all is well.)2012-07-15
  • 0
    @did Well, I wrote this answer because I found that most answers here depended on the derivative formula $D(e^x)=e^x$, but I thought it might as well be avoided because $D(e^x)=e^x$ is proved by this problem. $D(e^x)=e^x$ is what I really meant the property of $e^x$, which is not elementary.2012-07-15
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$\lim_{h\to 0}\frac{3^h-1}{h}=\lim_{h\to 0}\frac{e^{h\log 3}-1}{h}$. 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 $a = 3$ $$\lim_{h \to 0} \frac {3^h-1} {h}=\ln 3.$$

  • 0
    -1. This precludes the desired result.2012-07-16