Question: A function slowly varying at infinity $L:(0,\infty)\rightarrow (0.+\infty)$, has the property that for any $\delta>0$ if $x$ is large enough, then $L(x)\leq x^\delta$. Does this property implies that there exists a positive constant $C$ (possibly depending on $\delta$) such that $L(x)\leq C\, x^\delta$ for any $x\geq 1$?

Answer: 
No, $\frac{1}{|\ln x |}$ is a function slowly varying at infinity, but it is unbounded around $x=1$.
As indicated by kmitov's answer, you need an additional condition: $L$ is bounded on $[1,X]$ for all $X>1$.

Given $\delta>0$ let $X\_\delta>1$ be such that $L(x)\le x^\delta$ if $x\ge X\_\delta$. Let $M\_\delta$ be such that $L(x)\le M\_\delta$ if $1\le x\le X\_\delta$. Then
$$
L(x)\le \max(1,M\_\delta)\,x^\delta.
$$