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?

  • $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    Apr 12, 2020 at 1:46

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – anatoly
    Apr 11, 2020 at 22:39
  • $\begingroup$ @anatoly: if that is true, then the conclusion you asked about holds. $\endgroup$
    – Clayton
    Apr 11, 2020 at 23:35

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