Q:

Does ${\frac{f(n)}{g(n)}} \rightarrow l$ imply ${\frac{f(n)}{g(n+1)}} \rightarrow l$?

Let us assume that the sequence ${\frac{f(n)}{g(n)}} \rightarrow l$. That is, the sequence converges to the limit $l$ where $f, g$ are functions that take in natural numbers.
Will it be true that the sequence ${\frac{f(n)}{g(n+1)}}$ will also converge to the same limit $l$?

A:

Suppose $f(n) = 2^n$ and $g(n) = 2^{n+1}.$ Then $\dfrac{f(n)}{g(n)} = \dfrac 1 2$ for every value of $n$ and $\dfrac{f(n)}{g(n+1)} = \dfrac 1 4$ for every value of $n.$

