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?
 A: First, note that your problem seems to have two typos. As written, for $x\leq 0$ you get
$$|f(x)|<x\epsilon\leq 0$$
which is clearly impossible. It only makes sense if we change this inequality to 
$$|f(x)|\leq |x|\epsilon$$
You have the right idea for part $a)$, but your proof is lacking detail. Basically, when proving continuity, you need to use the fact that $|x|<\delta$ somewhere in the proof. This doesn't appear anywhere in your proof. I'll work 'backwards' and then show how this leads to a more rigorous proof. We want 
$$|f(x)-f(0)|<\epsilon$$
Further, let $\delta$ be defined by the problem (the existence portion). That is, let $|x|<\delta$. Separate with the triangle inequality (this is the backwards step) to get
$$|f(x)|+|f(0)|<\epsilon$$
Now, replace $|f(0)|$ and $|f(x)|$ with the condition given to get
$$|f(x)|+|f(0)|\leq |x|\epsilon +0=|x|\epsilon<\epsilon$$
This is true if $|x|<1$. Now, we have all the pieces here for a formal proof. Let $\epsilon>0$ be given and let $\delta$ be defined from the problem. Define
$$\Delta=\min\left\{1,\delta\right\}$$
Then $0<|x|<\Delta$ implies
$$|f(x)-f(0)|\leq |f(x)|+|f(0)|\leq |x|\epsilon +0\cdot \epsilon =|x|\epsilon<\epsilon$$
We conclude that $f(x)$ is continuous at $f(0)$.
For question $b)$, there is a similar difficulty with bad definitions. Ignoring the problem presented by $x\leq 0$, consider the function $f(x)=\frac{1}{\sqrt{x}}$. Then for any $\epsilon>0$, note that $0<x<\delta$ where $\delta$ is defined by
$$\delta=\epsilon^2$$
gives
$$x<\delta=\epsilon^2$$
$$\sqrt{x}<\epsilon$$
$$f(x)=\frac{1}{\sqrt{x}}<\frac{\epsilon}{x}$$
Thus, $f(x)$ satisfies the inequality given but clearly
$$\lim_{x\to 0^{+}}f(x)=\infty$$
Thus, $b)$ is false if one glosses over the difficulties arising from the definitions given.
A: 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.
A: The problem of the solution (a) is that $\frac {\epsilon + f(0)} {\epsilon} = 1 + \frac {f(0)} {\epsilon}$. As we don't know what value $f(0)$ is, we can't say that $x$ can become arbitrarily near 0; maybe $f(0)$ is some positive number, which will cause $x$ to approach infinity as $\epsilon \to 0$.
Part (b) is the same; maybe $f(0)$ is just 0 and $x$ will always be $1$; which serves no purpose.
