Arithmetic doubt while studying limits of sequences While studying limits of sequences, I came across these expressions.
$\left|\frac{-5}{n+2}\right|<\delta \iff\frac{5}{n+2}<\delta \iff n+2>\frac{1}{\delta }$
$n\in \mathbb N$
$\delta\in \mathbb R$ and $\delta >0$
I'm struggling to understand how it went from the second expression ($\frac{5}{n+2}<\delta$) to the third ($n+2>\frac{1}{\delta }$)
 A: Firstly, I think it shouldn't be an = because that's the confusing bit, it should should be an implication sign.
If that is the case then,
$\lvert\frac{−5}{n+2}\rvert< δ => \frac{5}{n+2} < δ => 5 < δ(n+2)$ 
$=> \frac{5}{δ} < n+2$
Since $\frac{1}{δ} < \frac{5}{δ}, => \frac{1}{δ} < n+2$
This is assuming that $n+2 > 0$ and $δ > 0$.
Hope this answers your question!
A: I guess we are assuming $n$ is positive, and as stated in the comments, this is probably a good place to use the implication sign, meaning that 
$$\frac{5}{n+2}<\delta\quad  \text{implies, or } \implies n+2>\frac{1}{\delta}$$
And this is because we can take the reciprocal of each side of your second inequality (and remember to flip the direction!) to get
$$\frac{n+2}{5}>\frac{1}{\delta}$$
And for positive $x$, we have $x>y \implies x+4x > y \implies5x>y $ so we have
$$n+2>\frac{1}{\delta}$$
A: To treat that as an equal sign just makes it dead wrong.  What they mean is implies the next.   (Actually; we must assume $\delta > 0$ and that $n + 2 > 0$.  I presume those are conditions stated in the source that you omitted?  That $n\in \mathbb N$?)
(Note: these implications only work one way.)
$\left|\frac{-5}{n+2}\right|<\delta \implies\frac{5}{n+2}<\delta \implies n+2>\frac{1}{\delta }$
$|\frac {-5}{n+2}| < \delta \implies \frac {5}{n+2} < \delta$.
That should be clear $|\frac {-5}{n+2}|= \frac {5}{n+2}$ (if $n+2 > 0$).  We could make the stronger.  $|\frac {-5}{n+2}| < \delta \iff \frac {5}{n+2} < \delta$.
Now...
$\frac{5}{n+2}<\delta \implies n+2>\frac{1}{\delta }$
This is presumably the part you have trouble with.
$\frac{5}{n+2}<\delta \iff$
$\frac{5}{n+2}\frac{n+2}{\delta} < \delta \frac{n+2}{\delta}\iff$ (assuming $n+2 > 0; \delta > 0$)
$\frac 5{\delta} < n+2$
So $\left|\frac{-5}{n+2}\right|<\delta \iff\frac{5}{n+2}<\delta \iff n+2>\frac{5}{\delta }$
Now $\frac 1{\delta} < \frac 5{\delta}$ so
$\frac 1{\delta} < \frac 5{\delta}< n+2$.
So $\frac {5}{n+2} \le \delta \implies n+2 > \frac 1{\delta}$
But obviously that is only one way.  $n+2 > \frac 1{\delta} \implies (\frac 1{n+2} < \delta; \frac 1{\delta} < \frac 5{\delta}) \not \implies \frac {5}{n+2} < \delta$.
A: In order to refer to them more easily, let’s label the inequalities as follows:
a) $\left\lvert\frac{-5}{n+2}\right\rvert < \delta$
b) $\frac{5}{n+2} < \delta$
c) $n+2>\frac 1\delta$
The fact that $n\in\mathbb N$ implies that $n+2>0$, in which case 
$\left\lvert\frac{-5}{n+2}\right\rvert = \frac{5}{n+2}$
and therefore a) and b) actually are equivalent. 
Again, since $n+2>0$ and $\delta>0,$ then 
$\frac{n+2}{\delta}>0$
and we can multiply both sides of b) by 
$\frac{n+2}{\delta} $ 
while preserving the direction of the $<$ sign.
The result is
$\frac{5}{\delta}<n+2.$
Now just turn that around:
$$ n+2 > \frac{5}{\delta}.$$
The inequality above is exactly equivalent to b), since we have just derived it from b) and can multiply it by 
$\frac{\delta}{n+2}$ to get b) again. 
It should be obvious enough that 
$\frac 5\delta > \frac 1\delta,$
and therefore b) implies c).
But consider the case $n=1,$ $\delta=1.$ In that case c) is true but b) is false. 
Therefore the two statements are not equivalent, and if your book says they are it is an error. 
