How to bound $\left\vert{\frac{x+2}{x+3}}\right\vert$ given that $1\lt x \lt 3$.

Question

I'm looking to bound $$\left\vert{\frac{x+2}{x+3}}\right\vert$$ given that $$1\lt x \lt 3$$. I would like to have my reasoning checked please.

My reasoning is as follows:

$$1\lt x \lt 3\implies (3\lt x+2\lt 5)$$ and $$(4\lt x+3\lt 6)$$.

Now consider $${\frac{x+2}{x+3}}$$:

as $4\lt x+3 \lt 6$ it follows that $$\frac{x+2}{6}\lt \frac{x+2}{x+3}\lt \frac{x+2}{4} $$.

And as $3\lt x+2\lt 5$, it must be that:

$$\frac{3}{6}\lt\frac{x+2}{6}\lt \frac{x+2}{x+3}\lt \frac{x+2}{4} \lt \frac{5}{4} $$.

Hence $$0\lt \frac{x+2}{x+3} \lt \frac{5}{4} $$.

Which implies that $$\left\vert{\frac{x+2}{x+3}}\right\vert \lt \frac{5}{4}$$.

(The reason I ask is that the problem set I'm working through had a question that required us to show that $$\frac{3}{4}\lt \frac{x+2}{x+3} \lt \frac{5}{6}$$ for $$1\lt x \lt 3$$ It then uses a sequences of equivalences to go from $$\frac{3}{4}\lt \frac{x+2}{x+3} \lt \frac{5}{6}$$ to $$1\lt x \lt 3$$ ) But I am interested in starting from $$1\lt x \lt 3$$ and then deducing a bound for the rational function. Is my reasoning correct?

Also, why does my reasoning lead to a less strict bound than the prescribed method?

Answer 1

$$\dfrac{x+2}{x+3}=1-\dfrac1{x+3}$$

$$1<x<3\iff4<x+3<6\iff\dfrac16<\dfrac1{x+3}<\dfrac14$$

$$\iff1-\dfrac16>1-\dfrac1{x+3}>1-\dfrac14$$

Answer 2

Your reasoning looks correct.

You could alternatively argue that the rational function $f(x)=\dfrac{x+2}{x+3}$ is continuous

and $\dfrac {df}{dx}=\dfrac1{(x+3)^2}$ is always positive,

so the bounds are at the end points $x=1$ and $3$, i.e., $f(1)=\dfrac34$ and $f(3)=\dfrac56$.

Answer 3

By $y=x+3$ the problem is equivalent to find the bounds for $4<y<6$

$$\left|\frac {y-1}{y}\right|=1-\frac1y$$

Answer 4

Putting $f(x)=\frac{x+2}{x+3}$, we get $f^{\prime}(x)=\frac{1}{(x+3)^{2}}$ which is a positive quantity ,hence we have $f(1)<f(x) <f(3)$ or $\frac{3}{4}<\frac{x+2}{x+3}<\frac{5}{6}$

Answer 5

Since $$\frac{x+2}{x+3}=1-\frac{1}{x+3},$$ we see that $\frac{x+2}{x+3}$ increases.

Thus, $$\frac{3}{4}=\frac{1+2}{1+3}<\frac{x+2}{x+3}<\frac{3+2}{3+3}=\frac{5}{6}.$$