We are trying to show that $\forall \epsilon > 0 \space \exists \delta>0 \space 0<|x-1|<\delta \Rightarrow |log(x)| < \epsilon$
My attempt:
Notice that if $|\log(x)| < \epsilon \Rightarrow -\epsilon < \log(x) < \epsilon \Rightarrow e^{-\epsilon} - 1 < x-1 < e^{\epsilon} - 1$.
Proof.
Take $\delta = e^{\epsilon} - 1$. We know that $0<|x-1|<\delta \Rightarrow x-1 < \delta \Rightarrow x < \delta + 1$. Now consider $|f(x)| = |\log(x)| < \log(\delta + 1) = \log(e^{\epsilon} + 1 - 1) = \epsilon $, as required.
Is this correct? I feel like I'm missing something, since $e^{-\epsilon} - 1$ can also be $\delta$.
Thanks.
Method 1
Set $x=e^u$. When $x\to 0$, then $u\to 0$, and thus $$\lim_{x\to 1}\log(x)=\lim_{u\to 0}\log(e^{u})=\lim_{u\to 0}u=0.$$
Method 2
$\log$ can be defined as $$\log(x)=\int_1^x\frac{1}{t}dt.$$ By the fundamental theorem of analysis, $$x\longmapsto \int_1^x\frac{1}{t}dt$$ is continuous on $(0,\infty)$, and thus $$\lim_{x\to 1}\log(x)=0.$$