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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$$\lim_{x\to 0} \frac{\log_a(1+x)}{x} = \lim_{x\to 0} \frac{\log_e(1+x)}{x\log_e{a}}=\lim_{x\to0} \frac{\sum_{i=1}^{\infty} (-1)^{n+1} \frac{x^{n}}{n}}{x\log{a}} =\lim_{x\to 0} \frac{1+\sum_{i=2}^{\infty} (-1)^{n+1} \frac{x^{n}}{n}}{\log{a}} = \frac{1}{\log{a}} = \frac{1}{\frac{\log_a{a}}{\log_{a}{e}}} = \log_{a}{e}$$
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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$$\lim_{h\to 0}\frac{b^h - 1}{h}$$
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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Consider: $$\frac{d}{dx}x^x$$ $$=\frac{d}{dx}e^{\ln{x^x}} = \frac{d}{dx}e^{x\ln{x}}$$ $$=e^{x\ln{x}}(\ln{x} + x\frac{1}{x})$$ $$=x^x(\ln{x} + 1)$$
Also: $$\frac{d}{dx}x^x$$ $$=\lim_{h\to 0}\frac{(x+h)^{x+h} - x^x}{h}$$ $$=\lim_{h\to 0}\frac{x^{x+h}+(x+h)hx^{x+h-1}+O(h^2)-x^x}{h}$$ $$=\lim_{h\to 0}x^x (\frac{x^{h}}{h}+(x+h)x^{h-1}-\frac{1}{h})$$ $$=x^x \lim_{h\to 0}(\frac{x^{h}}{h}+x^h-\frac{1}{h})$$ $$=x^x \lim_{h\to 0}(\frac{x^{h}}{h}+1-\frac{1}{h})$$
Therefore: $$x^x(\ln{x} + 1) = x^x \lim_{h\to 0}(\frac{x^h}{h}+1-\frac{1}{h})$$ $$\ln{x} = \lim_{h\to 0}(\frac{x^h}{h}-\frac{1}{h})$$
Mike Jerred
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