Hi. I answered this question but I have no idea if I did it correctly...

Question: Find a $\displaystyle \delta > 0$ that fulfills: if $\displaystyle |x+1| < \delta$, then $\displaystyle \left| \frac{x}{2x+3} + 1 \right| < \frac{1}{100}$

Answer: $\displaystyle \left| \frac{x}{2x+3} + 1 \right|$ = $\displaystyle \left| \frac{3x+3}{2x+3} \right| = \frac{3|x+1|}{|2x+3|}$

That part was easy, now is the part I don't know if I did logically/correctly...

The function is not defined at $\displaystyle x = -\frac{3}{2}$ so we can require $\displaystyle 0 \leq \delta \leq \frac{1}{5}$ (so the x can stay away from $\displaystyle -\frac{3}{2}$)

so that $\displaystyle |x+1| < \frac{1}{5}$ $\displaystyle \rightarrow$ $\displaystyle - \frac{1}{5} \leq x+1 < \frac{1}{5}$ $\displaystyle \rightarrow$ $\displaystyle 0 < \frac{3}{5}< 2x+3 < \frac{7}{5}$$\displaystyle \rightarrow$$\displaystyle |2x+3|>\frac{3}{5}$

So now we get

$\displaystyle \frac{3|x+1|}{|2x+3|} \leq \frac{3|x+1|}{\frac{3}{5}} = 5|x+1|$

and since $\displaystyle |x+1| < \delta$, $\displaystyle 5|x+1|<5\delta$

Therefore if we require $\displaystyle 5\delta < \frac{1}{100}$, we will for sure get for $\displaystyle |x+1|< \delta$

$\displaystyle 5|x+1| < 5\delta < \frac{1}{100}$

and so $\displaystyle 0< \delta < \frac{1}{500}$

Thanks!

**edit thanks for your advice!