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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 $3|x-1|$.

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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 $\delta:=\min\{1,\frac{\epsilon}{2}\}$

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  • $\begingroup$ Thank you, I realise this now. $\endgroup$ – Benjamin Aug 1 '19 at 21:33
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$x < 2 \Rightarrow \big|\frac{6(x-1)}{x}\big|>\big|\frac{6(x-1)}{2}\big|$, not $x < 2 \Rightarrow \big|\frac{6(x-1)}{x}\big|<\big|\frac{6(x-1)}{2}\big|$

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  • $\begingroup$ I realise my error now $\endgroup$ – Benjamin Aug 1 '19 at 21:33
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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 $\delta<\varepsilon/6$.

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  • $\begingroup$ I realise the mistake $\endgroup$ – Benjamin Aug 1 '19 at 21:33

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