Lets consider any interval $[a, b]$ and function $f(x) = \ln(x)$. I want to choose nodes $a \le a_1 \le a_2\le...,\le a_n \le b$ so that $$L_n(a_1,...,a_n;f) \rightarrow f$$
where $L_n(a_1,...,a_n;f)$ stands for interpolation of function $f$ in nodes $a_1,...,a_n$ as:
$$L_nf(x) = \sum_{j = 0}^n f(a_j)\prod_{k = 0, k \ne j}^n \frac{x - a_k}{a_j - a_k}$$
My work so far
My idea was to choose Chebyshev nodes (since those are very often optimal on interval):
$$a_i = \frac 1 2 ( a + b) + \frac 1 2 (b - a)\cos(\frac{2i - 1}{2n})\pi$$
then we know that:
$$\|f - L_n(a_1,...,a_n)\| \le \frac{\|f^{(n + 1)}\|}{(n + 1)! 2^{n - 1}}(\frac{c - b}{2})^n$$
Let us observe that:
$f(x) = \ln x$, $f'(x) = \frac 1 x$, $f''(x) = (-1) x^{- 2}$, $f'''(x) = (-1) \cdot (-2) \cdot x^{-3}$.... So we can observe, that $$f^{(n + 1)}(x) = \frac{n!(-1)^n}{x^{n +1}}$$
and from this place, we see that
$$\|f^{(n + 1)}\| = \max_{[a, b]}|\frac{n! (-1)^n}{x^{n + 1}}| = \frac{n!}{a^{n + 1}}$$
Finally:
$$\|f - L_n(a_1,...,a_n) \| \le \frac{n!}{(n + 1)! 2^{n - 1}}(\frac{b - a}{2})^n \frac{1}{a^{n + 1}} = $$ $$= \frac{1}{n + 1} \cdot \frac{1}{2^{n - 1}}\cdot \frac{1}{a2^n} \cdot (\frac b a - 1)^n$$
And here I have a problem, in my opinion, the above expression doesn't have to converge to $0$. If we consider the situation, that $b$ is huge, and $a$ is small, then expression $(\frac{b}{a} - 1)^n$ will be huge and will converge to $\infty$ much faster than other terms. Am I correct? Do you know maybe, what are the optimal nodes in this situation?