Limit existence proof

Question

Suppose I have limit $$ \lim_{n\to \infty }\left(f(n)+g_1(n)\right)=k \tag{1} $$ And $$ \lim_{n\to \infty }\frac{g_1(n)}{g_2(n)}=1 \tag{2} $$

Can I conclude that following limit exists $$\lim_{n\to \infty }\left(f(n)+g_2(n)\right)=k$$

If so how can i prove this?

My attempt

Add (1) to (2) and use limit sum law $$ \lim_{n\to \infty }\left(f(n)+g_1(n)+\frac{g_1(n)}{g_2(n)}\right)=k+1 $$ $$ \lim_{n\to \infty }\left(f(n)+\frac{g_1(n) g_2(n)}{g_2(n)}+\frac{g_1(n)}{g_2(n)}\right)=k+1 $$ $$ \lim_{n\to \infty }\left(f(n)+1 g_2(n)\right)+1=k+1 $$ And finally $$ \lim_{n\to \infty }\left(f(n)+g_2(n)\right)=k $$ Is this correct?

See this answer for a deep understanding of limit laws rather than trying to develop specific scenarios and guess what steps are allowed and what not. — Paramanand Singh, Apr 12 '20 at 01:46

score 5 · Accepted Answer · answered Apr 11 '20 at 19:26

5

It isn't true. Take $f(n)=n^2$, $g_1(n)=-n^2+k$, and $g_2(n)=-n^2+n$. We have $$\lim_{n\to\infty}f(n)+g_1(n)=k$$ and $$\lim_{n\to\infty}\frac{g_1(n)}{g_2(n)}=1,$$ but $$\lim_{n\to\infty}f(n)+g_2(n)=\infty.$$

Your mistake lies in the second-to-last displayed equation where you take the limit of just one part of the second term... this implicitly assumes convergence of the other sequences when you don't necessarily have that convergence.

answered Apr 11 '20 at 19:26

Clayton

24,751

Thanks. As far as I have understood, if $$\lim_{n\to \infty }\left(g_1(n)-g_2(n)\right)=0$$ then it works, right? – anatoly Apr 11 '20 at 22:39
@anatoly: if that is true, then the conclusion you asked about holds. – Clayton Apr 11 '20 at 23:35

Limit existence proof

1 Answers1