How to prove $\lim_{x\to1} \log(x)=0$ in a fair way? I am trying to prove the following limit with the definition:
$$\lim_{x\to1} \log(x)=0$$
That is - I must prove that for all $\epsilon>0$, there exists $\delta >0$ such that:
$$ 0<|x-1|< \delta \implies |\log x - 0|< \epsilon$$
So - my trouble is to understand what is "fair" to use here. I know the following:

*

*$x<e^x \implies \log(x)<x$
And it would be nice if we could find that

*

*$\log(x)\leq x-1$
From this I think we get:
$$|\log(x)|\leq|x-1| \text{ for }x\in[1,\infty]\qquad |\log(x)|\geq|x-1| \text{ for }x\in[-\infty,1]\tag{$\star$}$$
The trouble for me is that I know $\log(x)\leq x-1$ is true because of two things:

*

*Because $\log(1)=1-1=0$ and hence as $\log(x)$ and $x-1$ start in the same point and $\log'(x)=\frac{1}{x}$ and $(x-1)'=1$ then $\log (x)$ increases slower than $x-1$ when $x>1$ and decreases faster than $x-1$ when $x<1$. But is it fair to use derivatives? In the book I am reading, we didn't even reached derivatives yet.


*I looked a plot of both functions, which also doesn't seems fair.
Also, is it fair to assume we know that $\log(1)=0$? For me, knowing this seems to defeat the purpose of the problem somehow and the inequalities  I obtained in $(\star)$, the first would be helpful but I don't see how the second would help me.
So how can we proceed to prove this in a "fair" way?
 A: I guess the fairness depends on your definition of $\log(x)$.
Version 1 (hystorical). We may define $\log(x)$ as $\int_{1}^{x}\frac{dt}{t}$ (essentially the work in a isothermal process) and $e^x$ as the inverse function of $\log(x)$. By this way we have that $\log(x)$ is differentiable, increasing, concave and fulfills $\log(xy)=\log(x)+\log(y)$, so the inverse function $\exp(x)$ is differentiable, increasing, convex and fulfills $\exp(x+y)=\exp(x)\exp(y)$.
Version 2 (current). We may define $\exp(x)$ as $\sum_{n\geq 0}\frac{x^n}{n!}$ or as the unique solution of the Cauchy problem $f'(x)=f(x)$ with $f(0)=1$. By this way we have that $e^x$ is differentiable, increasing, convex and fulfills $\exp(x+y)=\exp(x)\exp(y)$. By naming $\log(x)$ the inverse function of $\exp(x)$ we have that $\log(x)$ is differentiable, increasing, concave and fulfills $\log(xy)=\log(x)+\log(y)$.
Tomato, tomato: in the first version, for any $\varepsilon\in(-1,1)$ we have
$$ \log(1+\varepsilon)=\int_{0}^{\varepsilon}\frac{dt}{1+t}\Longrightarrow \left|\log(1+\varepsilon)\right|\leq \frac{|\varepsilon|}{1-|\varepsilon|}. $$
In the second version we have
$$ \exp\left(\lim_{z\to 1}\log(z)\right)=\lim_{z\to 1}z=1 $$
and in both cases $\lim_{z\to 1}\log(z)=0$.
A: One way to define $\ln x$ is as $$\int_1^x\dfrac1t\rm dt$$.
Then if we take the limit we get $$\lim_{x\to1}\ln x=\lim_{x\to1}\int_1^x\dfrac 1t\rm dt=\int_1^1\dfrac 1t\rm dt=0$$, if we know $\ln x$ is continuous at $1$.
But in fact,  by the fundamental theorem of calculus, it's differentiable there.
