I know this is probably simple, but I am stuck. I want to prove this limit from the formal definition of a limit. How would you choose $\delta$ to make this work?

Let f: $\mathbb{R}$\{-1} $\to$ $\mathbb{R}$ be defined by $f(x)=\frac{x}{x+1}$. I want to show that $\lim_{x\to 3}$ $f(x)=\frac{3}{4}$

Want: $\forall$ $\epsilon > 0$, $\exists \delta > 0$, s.t. $\forall$ x $\in$ dom[f], $\Big[$(0<|x-3|

Given $\epsilon > 0$

Choose $\delta = \cdots$

Given x $\in$ dom[f]

Want $\Big[$(0<|x-3|

Assume (0 < |x-3|< $\delta$)

Want (|f(x)-$\frac{3}{4}$| < $\epsilon$)

(after some algebra)

Want (|$\frac{x-3}{x+1}$|<4$\epsilon$)

How do we choose $\delta$ in terms of $\epsilon$ to make this work?

  • $\begingroup$ If $|x-3|<\delta$, then $x+1>4-\delta$. $\endgroup$ – Taylor Nov 18 '15 at 19:39

This is my attempt....

Let $\epsilon>0$. Find some $\delta >0$ such that $\lvert \frac{x}{x+1}-\frac{3}{3+1}\rvert=\lvert \frac{x}{x+1}-\frac{3}{4} \rvert =\lvert \frac{4x-3(x+1)}{4(x+1)} \lvert =\lvert \frac{x-3}{4(x+1)} \rvert < \epsilon$ for all x with $0<\lvert x-3 \rvert <\delta$

$\lvert \frac{x-3}{4(x+1-4)+16} \rvert \le \frac{x-3}{4\lvert x-3 \rvert +16}<\frac{\delta}{4\delta +16}<\epsilon$

Then $\delta <\epsilon(4\delta +16)$

$\delta -4\delta \epsilon<16\epsilon$



So set $\delta =$ min{$\frac{16\epsilon}{1-4\epsilon}$}


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