# 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 $$\lim_{n\to\infty}f(n)+g_2(n)=\infty.$$
• 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