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How is

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

evaluated?

  • 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

12

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}.

  • 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
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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!

  • 2
    How is *philology* (with phi-, not phy-) related to the question?2012-07-15
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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 $(a^x-1)/x=y/\log_a(1+y)\to\ln a$.

  • 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, $\lim_{h\to 0}\frac{3^h-1}{h}=\log 3$.

-1

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