Question about the definition of limit Let $\lim_{x \rightarrow 0} f(x)=2$ then exists $\epsilon>0: \forall x \in (0,\epsilon)$ then :
a) $|f(x)|<4$
b) $|f(x)|>4$
c) $|f(x)|<1$
d) $|f(x)|<5$
For the definition of limit: $\forall r>0$ exists $\epsilon >0: \forall x \in D \cap I(0,\epsilon)-{0}$ then $2-r<f(x)<2+r$
So the right answer is c)?
 A: No, c) is not correct. Take $f(x)=2$, for instance. You have $\lim_{x\to0}f(x)=2$, but you never have $|f(x)|<1$.
And b) is not correct for the same reason.
But both a) and d) are correct. Can you see why?
A: 
then 2−r<f(x)<2+r

So if $f=0.1$ then $2-0.1=1.9 < f(x) < 2+0.1=2.01$ so $|f(x)| > 1.9 > 1$, so c) is wrong.
......
This is a really bass-ackward question, but ... forget about the $r$ for the time being.  This is a weird question that isn't actually asking how close we can get but how far away we not allowed to get.
As $x \to 0$ then $f(x) \to 2$ so for any $\epsilon > 0$ then interval $(0, \epsilon)$ will always contain points that are arbitrarily close to $2$.
Some of the choices will claim that all the points are away from $2$.  Those choices are right out.  Other of the choices will claim there are points not all that close to $2$ but that there could be, and we can surmise that if a point is arbitrarily close to $2$ then it'd also be within this larger distance as well  That means we can find an $\epsilon$ where they are all this close (and possibly closer).
c) If all $x\in (0,\epsilon)$ mean that $|f(x)| < 1$ then that means all $f(x): -1 < f(x) < 1 < 2$ and so $|f(x) - 2| > 1$.  That is NOT arbitrarily close to $2$. Or more accurately, we can always find an $x \in (0,\epsilon)$ where that is false. c) is wrong.
b)  If all $x\in (0,\epsilon)$ means that $|f(x)| > 4$ then that means for all $f(x): f(x) < -4$ of that $f(x) > 4$.  If $f(x) < -4 < 2$ then $|f(x)-2| > 6$.  If $f(x) > 4 > 2$ then $|f(x) - 2| > 2$.  In any event we can always find an $x \in (0,\epsilon)$ where that is false.  So b) is wrong.
a) If all $x \in (0,\epsilon)$ means that $|f(x)|<4$ then $-4 < f(x) < 4$ and that is allows for $-4 < 2-r < f(x) < 2+r < 4$ for $r$ small enough.
We can for instance:  If $r=2$ then we can find an $\epsilon$ where for all $x\in(0-\epsilon, 0+\epsilon)$ we have $|f(x) -2| < 2$.  So $-2 < f(x) -2 < 2$ so $0 < f(x) < 4$ so $|f(x)| < 4$.  And if $x \in (0, \epsilon)\subsetneq (0-\epsilon, 0+\epsilon)$ the result will hold.
So a) is true.
But we can also:  If $r = 0.01$ then we can find an $\epsilon'$ so that for all $x \in (-\epsilon' , \epsilon')$ then $|f(x)-2| < r$ so $1.99 < 2-r < f(x) < 2+r= 2.01$ and if $1.99 < f(x) < 2.01$ then $0 < 1.99< |f(x)| < 2.01 < 4< 5$.
So we can find such an epsilon that goes in even tighter.
d) ditto above.
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that's my loooooong analysis.
to actually do the problem I'd just do.
For any $r$ we can find an $\epsilon_r > 0$ so that for all $x \in (0,\epsilon)\subset (-\epsilon, \epsilon)$ so that $|f(x) - 2| < r$
That means $2-r < f(x) < 2+r$ and that means $-2-r < 2-r< f(x) < 2+r$ so $2-r< |f(x)| < 2+r$.
If we had chosen our $r$ so that $r<1$ then the $\epsilon$ we found would be so that $x\in (0,\epsilon)\implies$:  $1 < |f(x)|< 3$ and a) and c) are  false and b) and d) are true.
The end.
A: By defintion of the limit
$$\forall r>0\;\; \exists \epsilon>0\;\;:\;\;$$
$$ \forall x\in D\cap (-\epsilon,\epsilon)-\{0\}\;\;$$
$$ 2-r<f(x)<2+r$$
So, if $ r=2$ then
$$0<f(x)<4 \implies \;|f(x)<4$$
And if $ r=3$ then
$$-5<-1<f(x)<5\;\implies |f(x)|<5$$
$a$ and $d$ are right answers
A: Suppose that $f(x)=2$ for all $x$ in the domain, c) is obviously wrong.
