3
$\begingroup$

Prove the limit using definition of limit$$ \ \lim_{a \to \infty}\ \frac{3a+2}{5a+4}= \frac{3}{5} $$

Answer: Let $\varepsilon \ >0 $. We want to obtain the inequality $$\left|\frac{3a+2}{5a+4}- \frac{3}{5}\right|< {\varepsilon}$$
$$ \Rightarrow \left|\frac{3a+2}{5a+4} - \frac{3}{5}\right|\ =\left|\frac{5(3a+2)-3(5a+4)}{5(5a+4)}\right|\\= \left|\frac{-2}{5(5a+4)}\right|\le\frac{1}{a} $$ Therefore we choose $K \in N$ s.t $K> \frac{1}{\varepsilon} $ $$\Rightarrow\left|\frac{3a+2}{5a+4}- \frac{3}{5}\right| \le\frac{1}{a}\le\frac{1}{K} < \varepsilon $$ Is this correct?

$\endgroup$
4
  • $\begingroup$ You can use $\Rightarrow$ $\Rightarrow$ or $\implies$. It looks better than $=>$. $\endgroup$
    – Martin Sleziak
    Sep 26, 2015 at 7:32
  • 1
    $\begingroup$ The first sentence of your proof (in the current version of your post is): Choose $\varepsilon>0$ such that $k>\frac1\varepsilon$. Did you mean it other way round? Suppose we are given $\varepsilon>0$. Choose a positive integer $k$ such that $k>\frac1\varepsilon$. (If you are proving something using definition of limit, you should show that for *every $\varepsilon>0$ holds....) $\endgroup$
    – Martin Sleziak
    Sep 26, 2015 at 7:35
  • $\begingroup$ Martin Sleziak is right. And I also think you should add "when ever $a \geq k$ we have ..." $\endgroup$
    – R_D
    Sep 26, 2015 at 7:38
  • $\begingroup$ @MartinSleziak i just edited my post again. I think it is clearer now $\endgroup$
    – ross999
    Sep 26, 2015 at 7:48

2 Answers 2

Reset to default
3
$\begingroup$

Your proof seems fine to me. It seems clear from what you wrote that you know what you are doing. And I like that at the beginning of your proof you clearly stated what is your goal.

Still they are a few minor points. (You may call this nitpicking if you wish.)

  • The symbol $\Rightarrow$ is used to denote impication. I do not like the place where you wrote $$\Rightarrow \left|\frac{3a+2}{5a+4} - \frac{3}{5}\right|\ = \dots$$ because what you wrote here is in fact not a consequence of what you wrote previously. This is the place where you are trying to simplify the expression you need to estimate. (So I would simply omit $\Rightarrow$ at that place.)
  • The inequality $$\left|\frac{-2}{5(5a+4)}\right|\le\frac{1}{a}$$ is more-or-less clear, but you might add an explanation why this inequality holds. (I'd say this depends on who is grading this, if it is an assignment. Different people might expect different level of detail. The most important thing is that you understand why the inequality used in your proof holds.)

Since it is possible that you will further edit your post, I will add a link to the revision for which my comments were intended.

$\endgroup$
1
$\begingroup$

let $ζ>0$ be any arbitrary real number consider, $$\begin{align} &|[(3a+2)/(5a+4)]-(3/5)| < ζ\\ \Leftrightarrow&|-2/5(5a+4)| < ζ\\ \Leftrightarrow& a > 1/5[(2/5ζ)-4]=∆ \end{align}$$ hence for every $ζ>0$ there exists $∆>0$ such that which satisfies the definition of convergence hence $f(a)=[(3a+2)/(5a+4)]$ converges to $3/5$.

$\endgroup$
1
  • $\begingroup$ I have tried to edit your post for better readability. You may edit it further. (For example, you might write $\frac{3a+2}{5a+4}$, which is entered as $\frac{3a+2}{5a+4}$ if you think it looks better.) For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$
    – Martin Sleziak
    Sep 26, 2015 at 9:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .