Arithmetic doubt while studying limits of sequences

Question

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 ($n+2>\frac{1}{\delta }$)

Answer 1

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 $$n+2>\frac{1}{\delta}$$

Answer 2

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!

Answer 3

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. $n+2 > \frac 1{\delta} \implies (\frac 1{n+2} < \delta; \frac 1{\delta} < \frac 5{\delta}) \not \implies \frac {5}{n+2} < \delta$.

Answer 4

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.