Replacing the limsup and liminf of a function by using a sequence with bounded ratio of consecutive-terms

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Let $f:(0,\infty) \to \mathbb R$ be a continuous function and consider the $\liminf_{x\to 0} f(x)$ and $\limsup_{x\to 0} f(x)$. Let $x_n \to 0$ be any sequence satisfying that $\{x_n/x_{n+1} \}$ is bounded. I wonder if the following equalities are true and how to prove them:

$$\liminf_{x\to 0} f(x)=\liminf_{n\to \infty} f(x_n), \limsup_{x\to 0} f(x)=\limsup_{n\to \infty} f(x_n).$$

Clearly I have the trivial inequalities by the definitions of liminf and linmsup (being the smallest/largest sequencial limits)