# Limit existence proof

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. Apr 12, 2020 at 1:46

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.

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