How is
$$\lim_{h \to 0} \frac {3^h-1} {h}=\ln3$$
evaluated?
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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 $f'(0) = \log{3}$. |
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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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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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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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$\lim_{h\to 0}\frac{3^h-1}{h}=\lim_{h\to 0}\frac{e^{h\log 3}-1}{h}$. 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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