Proof that lim x->0 log(1+x)/x = 1

[uman]

Given that $$\log{(1+x)}=\int_0^x\frac{dt}{1+t}$$, how would one prove that $$\lim_{x \to 0}\frac{\log{(1+x)}}{x}=1$$?

 

[Manchot]

Just use l'Hôpital's rule and the fundamental theorem of calculus.

 

[uman]

Thanks! I had thought of that. However, L'Hôpital's rule is introduced much later in this book so I assume I'm not supposed to use it. It's so far covered logarithms, differentiation and integration techniques... any other ideas?

 

[Manchot]

Ok, if you can't use l'Hôpital, how about this? Subtract a log(1) from the numerator in the expression of the limit. It should look quite familiar now. :-D

 

[uman]

Sorry, I should have mentioned something: You're supposed to prove it using the identity I gave in the first part of my post. This is part B of an exercise in apostol. Part A was to prove it using your method, which I did.

The statement of part A was something like "Prove the relationship using the definition of f'(1), where f(x) = log(x)" or something of that nature.

 

[D H]

Express $1/(1+t)$ as an infinite series, integrate from 0 to x, divide by x, and take the limit as $x\to 0$.

 

[uman]

Meh. The book hasn't covered series' either. Oh well.

 

[Gib Z]

I haven't tried this yet, normally I would go with either of Monchot's suggestions, but you could try expressing the integral in terms of Riemann sums. I think that should work.

 

[Rainbow Child]

Given that $$\log{(1+x)}=\int_0^x\frac{dt}{1+t}$$, how would one prove that $$\lim_{x \to 0}\frac{\log{(1+x)}}{x}=1$$?

For the expression

$$I=\frac{1}{x}\,\int_0^x\frac{d\,t}{1+t}$$

apply the change of variables $t=x\,u,d\,t=x\,du$, so

$$I=\int_0^1\frac{d\,u}{1+x\,u}$$

Then

$$\lim_{x \rightarrow 0}I=\int_0^1 \lim_{x \rightarrow 0}\left(\frac{1}{1+x\,u}\right)\,d\,u=\int_0^1 d\,u=1$$.

You are allowed to change order between the integral and the limit, since the integral exists by hypothesis.

 

[Gib Z]

Rainbow Child's suggestion crushes mine :( But I just tried it, and it works as well =]

 

[uman]

Thank you Rainbow Child! I understand your proof and it makes perfect sense. Very clever!

 

[uppaladhadium]

Lim x->0 (x(1+(x/2)+((x^2)/3)+....))/x
=lim x->0 (1+(x/2)+((x^2)/3)+....)
=1+0+0+0......=1

 

[HallsofIvy]

As lovely as it is, Rainbow Child's suggestion requires knowing that, in this case,
$$\lim_{x\to 0} \int_0^1 \frac{dt}{1+ xt}= \int_0^1 \lim_{x\to 0}\frac{dt}{1+ xt}$$
which, while true, probably requires deeper math than uman, who has not yet dealt with L'Hopital's rule or power series, has available.