Is there a formal proof of this fact without using L'Hôpital's rule? I was thinking about using a proof of this fact: $$ \left.\frac{d(e^{x})}{dx}\right|_{x=x_0} = e^{x_0}\lim_{h\to 0} \frac{e^{h}-1}{h} = e^{x_0}\cdot 1=e^{x_0} $$ that I have to help prove: $$ \lim_{h\to 0} \frac{b^{h}-1}{h} = \ln{b} $$ Is there a succinct proof of this limit? How does one prove this rigorously?
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Careful @arbautjc, I believe you need the derivative of ln to do that, which requires evaluating the limit in question. – Dylan Yott Mar 15 '13 at 11:44
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I think you need only the derivative of $\exp$ (but, even if we need derivative of $\ln$, is it forbidden ?). Anyway, I removed my answer since it involves L'Hopital's rule and Rustyn explicitly wrote without L'Hopital's rule. – Jean-Claude Arbaut Mar 15 '13 at 11:50
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It is easy to see that $$\lim_{x\to 0}\frac{\log_a(1+x)}{x}=\log_a e$$ Now set $y=a^x-1$. So $a^x=1+y$ and then $$x=\log_a(1+y)$$ Clearly, when $x\to0$; $y\to 0$ so $$\lim_{x\to 0}\frac{a^x-1}{x}=\lim_{y\to 0}\frac{y}{\log_a(1+y)}=\log_e a$$
Mikasa
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By easy, do you mean using the Taylor series for $\log_{a}(1+x)$? – Robert Abramovic Nov 20 '23 at 17:36
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$$\text{ As }b=e^{\ln_eb}, \frac{b^h-1}h=\frac{e^{h\ln b}-1}h=\ln b\frac{e^{h\ln b}-1}{h\ln b }$$
$$\text{As, }\lim_{x\to 0}\frac{e^x-1}x=1,$$ $$\lim_{h\to0}\frac{b^h-1}h=\ln b\lim_{h\to0}\frac{e^{h\ln b}-1}{h\ln b }=\ln b$$
apnorton
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lab bhattacharjee
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You used the solution in the proof, that's a circular argument – Allan Henriques Oct 02 '20 at 19:55
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1No @AllanHenriques: they used the special case $b=e$, i.e. $\exp'(0)=1$. – Anne Bauval Jul 12 '24 at 04:39
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1@AnneBauval you're right. Been a while, I'm about to graduate with a degree in math and I was probably in first year around that point – Allan Henriques Jul 18 '24 at 02:28
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$$\lim_{x\to 0} \frac{\log_a(1+x)}{x} = \lim_{x\to 0} \frac{\log_e(1+x)}{x\log_e{a}}=\lim_{x\to0} \frac{\sum_{i=1}^{\infty} (-1)^{n+1} \frac{x^{n}}{n}}{x\log{a}} =\lim_{x\to 0} \frac{1+\sum_{i=2}^{\infty} (-1)^{n+1} \frac{x^{n}}{n}}{\log{a}} = \frac{1}{\log{a}} = \frac{1}{\frac{\log_a{a}}{\log_{a}{e}}} = \log_{a}{e}$$
Now set $y=a^x -1.$ Hence $a^x = 1+y$ and
$$
x = \log_a(1+y)
$$
As $x\to 0; y\to 0$
Therefore:
$$
\lim_{x\to 0}\frac{a^x-1}{x}=\lim_{y\to 0}\frac{y}{\log_a(1+y)}=\log_e a
$$
Thanks Babak.
Rustyn
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$$\lim_{h\to 0}{b^h-1\over h}$$ First, we define a variable t such that $b^h-1=t$. Therefore, $b^h=t+1$ and $h=\log_b{(t+1)}$. It should be noted that, if we take the limit as $h$ goes to $0$, $t$ also goes to $0$ as $b^0-1=0$. So, we redefine our original limit using $t$. $$\lim_{t\to 0}{t \over \log_b(t+1)}$$ Next, we move the $t$ in the numerator to the denominator. $$\lim_{t\to 0}{1 \over {1\over t}\log_b(t+1)}$$ Then, using properties of logarithms, $$\lim_{t\to 0}{1\over \log_b((t+1)^{1 \over t})}$$ Now let's take a look at the definition of $e$. $e$ is usually defined as $$e=\lim_{x\to\infty}{(1+{1\over x})^x}$$ However, if we define $r$ as the reciprocal of $x$, we see that as $x$ tends to infinity, $r$ tends to $0$. So, we can also write that $$e=\lim_{r\to0}{(1+r)^{1\over r}}$$ Notice that this is what we have inside of our logarithm in step 4. So, we can rewrite our limit as $${1 \over \log_b{e}}$$ We can rewrite 1 to get $${\log_b{b} \over \log_b{e}}$$ And finally, because of the change of base rule, $${\log_e{b}}$$
Credit to Massimiliano on Socratic.org for the logic of this post. (It's not an exact copy as he was not answering the exact same question, but they're very similar) Link here: How do you solve the limit (e^x-1)/x as x approaches 0?
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$$\lim_{h\to 0}\frac{b^h - 1}{h}$$
Let $x = \frac{b^h - 1}{h}$ $$h.x = b^h - 1$$ $$b^h = hx + 1$$
Applying $\ln$ both sides
$$h\ln{b} = \ln{(hx+1)}$$ $$\ln{b}=\frac{\ln{(h.x+1)}}{h}$$ $$\ln{b}=\frac{\ln{(h.x+1)}}{hx}.x$$ Now when $h\to0$;$hx\to0$;$\frac{\ln{(h.x+1)}}{hx}\to ln(e);x\to \ln(b)$
Thus we can say
$$\lim_{h\to 0}\frac{b^h - 1}{h} = \ln(b)$$
Abhijit
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