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!

  • 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
    Oct 29 '17 at 18:29
  • $\begingroup$ Would you show me how to start that prove? Or what would the procedure be? @Math1000 $\endgroup$
    – Evoked
    Oct 29 '17 at 19:07


Note that


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

  • $\begingroup$ 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
    Oct 29 '17 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
    Oct 29 '17 at 20:49

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.