7
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Careful @arbautjc, I believe you need the derivative of ln to do that, which requires evaluating the limit in question. $\endgroup$ – Dylan Yott Mar 15 '13 at 11:44
  • $\begingroup$ 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. $\endgroup$ – Jean-Claude Arbaut Mar 15 '13 at 11:50
5
$\begingroup$

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 $$\lim_{x\to 0}\frac{a^x-1}{x}=\lim_{y\to 0}\frac{y}{\log_a(1+y)}=\log_e a$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ good job, Babak! $\ddot\smile$ $\endgroup$ – amWhy Mar 16 '13 at 0:46
4
$\begingroup$

$$\text{ As }b=e^{\ln_eb}, \frac{b^h-1}h=\frac{e^{h\ln b}-1}h=\ln b\frac{e^{h\ln b}-1}{h\ln b }$$

$$\text{As, }\lim_{x\to 0}\frac{e^x-1}x=1,$$ $$\lim_{h\to0}\frac{b^h-1}h=\ln b\lim_{h\to0}\frac{e^{h\ln b}-1}{h\ln b }=\ln b$$

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

$$\lim_{x\to 0} \frac{\log_a(1+x)}{x} = \lim_{x\to 0} \frac{\log_e(1+x)}{x\log_e{a}}=\lim_{x\to0} \frac{\sum_{i=1}^{\infty} (-1)^{n+1} \frac{x^{n}}{n}}{x\log{a}} =\lim_{x\to 0} \frac{1+\sum_{i=2}^{\infty} (-1)^{n+1} \frac{x^{n}}{n}}{\log{a}} = \frac{1}{\log{a}} = \frac{1}{\frac{\log_a{a}}{\log_{a}{e}}} = \log_{a}{e}$$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You are Welcome. I hope I could help you. $\endgroup$ – mrs Mar 17 '13 at 7:41
0
$\begingroup$

$$\lim_{h\to 0}\frac{b^h - 1}{h}$$

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

$$\lim_{h\to 0}\frac{b^h - 1}{h} = \ln(b)$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.