1
$\begingroup$

I already know how to compute a limit by its definition when x tends to a real number. But know I questioned myself how can you prove a limit if it tends to infitity?. I looked for an exercise and found this:

$$\lim_{x\to\infty} = \frac{7x+2}{4x+3}-\frac{7}{4}$$

I tried to follow the normal path but got stucked in here:

$|\frac{13}{4(4x+3)}|<\epsilon ; |x-\infty|<\delta$

UPDATE: I kept operating and came up with: As $x\to+\infty;|\frac{13}{4(4x+3)}|=\frac{13}{4(4x+3)}$

And if $\frac{13}{4(4x+3)}<\epsilon\to\frac{4(4x+3)}{13}>\frac{1}{\epsilon}$

I think I'm near to the solution, but I fail on seeing what is the relation between $k$ and $\epsilon$

I would appreciate any help!

$\endgroup$
2
  • 1
    $\begingroup$ The definition of $\lim_{x\to\infty}f(x)=L$ is: For all $\varepsilon>0$, there exists $x_0$ such that $x\geqslant x_0$ implies $|f(x)-L|<\varepsilon$. $\endgroup$
    – Math1000
    Commented Oct 29, 2017 at 18:29
  • $\begingroup$ Would you show me how to start that prove? Or what would the procedure be? @Math1000 $\endgroup$
    – Evoked
    Commented Oct 29, 2017 at 19:07

1 Answer 1

1
$\begingroup$

HINT:

Note that

$$\left|\frac{7x+2}{4x+3}-\frac74\right|=\left|\frac{13}{4(4x+3)}\right|$$

Now show that for any $\epsilon>0$, there is a number $B>0$ such that $\left|\frac{13}{4(4x+3)}\right|<\epsilon$ whenever $x>B$.

$\endgroup$
2
  • $\begingroup$ I do not know how to show it, would you mind giving me a little help? Like from where do I start or something? Sorry, but I just still do not know how to compute the limit $\endgroup$
    – Evoked
    Commented Oct 29, 2017 at 18:46
  • $\begingroup$ Well, you've computed the limit already - $7/4$. Now, can you rearrange the inequality $\frac{13}{4(4x+3)}<\epsilon$ to isolate $x$? $\endgroup$
    – Mark Viola
    Commented Oct 29, 2017 at 20:49

You must log in to answer this question.

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