Let $\{x_n\}$ be a sequence converging to $L$. According to Wikipedia, if there exists a $\mu\in(0,1)$ satisfying
$$\lim_{k→∞}\frac{|x_{k+1}−L|}{|x_k−L|}=μ$$
then we say $\{x_n\}$ converges linearly to $L$ and has a rate of convergence of $\mu$. If $\mu = 0$ we say $\{x_n\}$ converges superlinearly and if $\mu = 1$ we say $\{x_n\}$ converges sublinearly.
My question is:
If $\{x_n\}$ converges to $L$, does the following hold? (See update below)
$$ \lim_{k\rightarrow\infty}|x_k|^{1/k} = \lim_{k→∞}\frac{|x_{k+1}−L|}{|x_k−L|}.$$
Intuitively these two limits seem like they should convey similar information about the speed of convergence, and trying a few simple sequences gave equivalent results.
Context:
This came up while studying Cramér's theorem of Large Deviations. Without getting into the specifics here, the theorem can be stated as the following limit:
$$ \lim_{n\rightarrow \infty} \operatorname{Pr}(M_n \geq x)^{1/n} = e^{-I(x)}$$
and the sources I am reading discuss (informally) that this result implies that the probability in the left hand side decays exponentially with $n$, but I do not understand why that is implied. My guess is that it is related to my question.
**Update: I was mistaken in my original question. As Fred pointed out, the equality was trivially false.
I am actually interested in whether the following holds:
$$ \lim_{k\rightarrow\infty}|x_k - L|^{1/k} = \lim_{k→∞}\frac{|x_{k+1}−L|}{|x_k−L|}.$$
Any guidance would be appreciated.
If $x_k \to L \ne 0$, then $|x_k| \to |L|>0$. Then there is $K$ such that
$\frac{1}{2}|L| \le x_k \le \frac{3}{2}|L|$ for all $k>K$.
Hence: $\lim_{k\rightarrow\infty}|x_k|^{1/k}=1$.