0
$\begingroup$

I know that if $$ \lim_{x\rightarrow\infty}(f(x)-a)=0 $$ for some a in the reals can I say $$\lim_{x\rightarrow\infty}f(x) =a$$

I know that the sum of limits existing does not imply that each limit must exist but in this case, since a is just a number I do not see how the limit of f(x) could not exist.

If so, how might I prove this better?

$\endgroup$
2
  • 1
    $\begingroup$ Yes, you are correct. Write down the definition of what it means for $f(x)-a$ to go to $0$ as $x\to \infty$ (using $\epsilon$ and $M$). It’s the same as the definition of what it means for $f(x)$ to go to $a$. $\endgroup$
    – User8128
    Dec 13, 2020 at 2:46
  • $\begingroup$ Yes you can provided that the limits of each term exist and are finite. If one term were going to positive infinity and the other to negative infinity, you have an indeterminate case. $\endgroup$
    – Doug M
    Dec 13, 2020 at 2:54

1 Answer 1

Reset to default
3
$\begingroup$

If $\lim_{x\to\infty}(f(x)-a)$ exists and is equal to $0$, then by the limit sum formula

$$\lim_{x\to\infty}f(x) = \lim_{x\to\infty}(f(x)-a) + \lim_{x\to\infty}a = a$$

$\endgroup$
2
  • $\begingroup$ Is this rigorous enough to be used? Or is more needed to show that f(x) does exist here. Just from looking, it seems like I can expand this so that $$a=mx+b$$ or some other simple polynomial (Of course then it would be equal to the limit if a is defined in terms of x). $\endgroup$
    – Derek C.
    Dec 13, 2020 at 2:54
  • 1
    $\begingroup$ @DerekC. The limit of sum rule requires that both $\lim_{x\to\infty} (f(x)-a)$ and $\lim_{x\to\infty} a$ exist. Then according to the rule, the limit $\lim_{x\to\infty}[(f(x)-a) - a]$ exists as a result, not as a prerequisite. $\endgroup$
    – peterwhy
    Dec 13, 2020 at 3:01

You must log in to answer this question.

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