We have given a function $f(x)$ where we know that $f(x)\leq 1$ for all $x$.
Is it true that $$1+O(f(x))=O(f(x))$$ even though I know that $1 \geq f(x)$?
We know that $O(f(x))=o(g(x))$. Is it true that $$1+O(f(x))=o(g(x))$$ for any $g(x)$?
Under which conditions does this hold?
Thank you for your help
Refer to the definitions. Consider some corner cases, e.g. $f(x) = 1 / x$, and you'll see that there is no way to find constants $N$ and $c$ such that $1 + 1 / x \le c / x$ whenever $x \ge N$.
As above, it is often useful to pick some carefully selected cases to explore the situation. Try to find counterexamples. If you find one, you are done; if you can't find one, the process might help in writing a proof.