# Why is $|x-x_0|^k$ decreasing in k as x gets closer to $x_0$?

This question related to the Taylor Series. The statement is: If f ∈ $$C^{k+1}$$ (A, R), then $$lim _{x->x_0}|R_k(x)/ (x-x_0)^{k+1}|< ∞$$.

1. Why do we not say $$R_{k+1}$$?
2. I read that as x gets closer to $$x_0$$, $$|x-x_0|^k$$ is going to decrease in k, hence the higher the k, the higher is the accuracy. What does this exactly mean? What does decreasing in k mean?
• 1. We say $R_k$ because it is the remainder if we take the taylor-expansion of degree n, this is always a degree higher than n. For 2. I could only guess, that if $|x-x_0|<0$ then increasing $k$ decreases the expression. Where did you read it? Maybe it becomes clear from the context. Commented Dec 30, 2022 at 21:43

When $$0<|x-x_0|<1$$, then $$|x-x_0|^k = \exp\big(\log |x-x_0| \cdot k \big).$$ Since $$\log |x-x_0|< \log(1) < 0$$, the function $$k \mapsto \log |x-x_0| \cdot k$$ is decreasing. Since $$z\mapsto \exp(z)$$ is a strictly increasing function, it preserves monotonicity, thus $$k \mapsto \exp\big(\log |x-x_0| \cdot k \big)$$ is decreasing.