1
$\begingroup$

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?

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

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

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

$\delta(1-4\epsilon)<16\epsilon$

$\delta<\frac{16\epsilon}{1-4\epsilon}$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.