Conditions that may make a function continuous at the origin 
ATTEMPT:
a) My answer is afirmative. Since for any $x$ we have the condition, then if we put $x = \dfrac{\epsilon + f(0) }{\epsilon} $, then
$$ |f(x)| < \epsilon + f(0) \implies |f(x)-f(0)| < \epsilon $$
and since $\epsilon$ is arbitrary then $f(x)$ is continous at $x=0$
b) Similarly, but this time, we can put $x = \dfrac{ \epsilon }{\epsilon + f(0)} $
Is this correct?

As @QC_QAOA said, the statement of the problem in (a) should be: Suppose that for every $\epsilon>0$ there exists $\delta > 0$ such that $|x|<\delta$ implies $|f(x)|\le \epsilon|x|.$ Is $f$ continuous at $0?$
The answer is yes. Here's a preliminary result: Suppose $a>0$ and $|f(x)|\le |x|$ for $|x|<a.$ Then $f$ is continuous at $0.$ To prove this, note we easily have $f(0)=0.$ Let $\epsilon>0.$ Choose $\delta =\min(a,\epsilon).$ Then $|x|<\delta$ implies
$$|f(x)-f(0)| = |f(x)|\le |x| <\delta = \epsilon.$$
To finish off (a), let $\epsilon =1.$ By hypothesis there is $a>0$ such that $|x|<a$ implies $|f(x)|\le 1\cdot |x|=|x|.$ We're done by the preliminary result.
As for (b), there is really no hope as far as I can see. To make sense out of $|f(x)|\le \epsilon/|x|$ for $|x|<\delta$ we either need to change that to $0<|x|<\delta,$ or else allow $x=0$ and declare $\epsilon/0=\infty.$ In either scenario, we could let $f(0)=1$ and $f=0$ everywhere else to get a counterexample.
