# Is this $\epsilon-\delta$ proof correct?

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 $$|x-1|<1 \implies 0. Hence $$\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 $$3|x-1|$$.

If $$|x-1|<1$$ then $$0 which implies $$|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}\}$$
$$x < 2 \Rightarrow \big|\frac{6(x-1)}{x}\big|>\big|\frac{6(x-1)}{2}\big|$$, not $$x < 2 \Rightarrow \big|\frac{6(x-1)}{x}\big|<\big|\frac{6(x-1)}{2}\big|$$
Well, $$|x|<2$$ implies $$1/|x|>2$$, and your conclusion that $$\left|\frac{6(x-1)}x\right|<|6\delta/2|$$ doesn't hold. Instead, notice that around $$x=1,x+3>1$$ so $$1/|x+3|<1$$. Thus,$$\left|\frac{6(x-1)}{x+3}\right|<|6(x-1)|<6\delta<\varepsilon$$giving $$\delta<\varepsilon/6$$.