I would like to verify whether my proof of the following limit is correct.
Show that $\lim_{x\to 1}\frac{8x}{x+3}=2$.
Rough work:
If $\delta<1$ then $|x-1|<1 \implies 0<x<2\implies |x|<2$. Hence $\big|\frac{8x}{x+3}-2\big| =\big| \frac{6(x-1)}{x+3}\big| < \big|\frac{6(x-1)}{x}\big|<\big|\frac{6(x-1)}{2}\big|< 3|x-1|$.
Proof: Given $\epsilon>0$, take $\delta:=\min\{1,\frac{\epsilon}{3}\}$. Then $|x-1|<\delta$ implies $$\big|\frac{8x}{x+3}-2\big|< \big|\frac{6(x-1)}{x}\big|<\big|\frac{6(x-1)}{2}\big|< 3|x-1|<3\delta=\epsilon.$$$$\tag*{$\blacksquare$}$$
In particular I want to verify that my rough work is correct and that my steps are valid in reaching the estimation $3|x-1|$.
Best Answer
If $|x-1|<1$ then $0<x<2$ which implies $|x+3|=x+3>3$ and
$$\left|\frac{8x}{x+3}-2\right| =\frac{6|x-1|}{|x+3|}< \frac{6|x-1|}{3}=2|x-1|.$$ Therefore it suffices to take $\delta:=\min\{1,\frac{\epsilon}{2}\}$