Limit Related Conceptual question If $\lim_{x\to\infty} \dfrac{f(x)}{g(x)} = 1$ and $\lim_{x\to\infty} f(x) = c$, must $\lim_{x\to\infty} g(x)$ necessarily be c? What about $\lim_{x\to\infty} g(x) = c$, must $\lim_{x\to\infty} f(x) = c$? Proof?
 A: We just need to use the rules dealing with algebra of limits. In this case following two rules are needed:

1) If $\lim_{x \to \infty}f(x)$ and $\lim_{x \to \infty}g(x)$ exist then $$\lim_{x \to \infty}f(x)g(x) = \lim_{x \to \infty}f(x)\cdot\lim_{x \to \infty}g(x)$$
2) If $\lim_{x \to \infty}f(x)$ and $\lim_{x \to \infty}g(x) \neq 0$ exist then $$\lim_{x \to \infty}\frac{f(x)}{g(x)} = \dfrac{{\displaystyle \lim_{x \to \infty}f(x)}}{{\displaystyle \lim_{x \to \infty}g(x)}}$$

Now we are given that $$\lim_{x \to \infty}\frac{f(x)}{g(x)} = 1\tag{1}$$ and $$\lim_{x \to \infty}f(x) = c\tag{2}$$ Then we have $$\lim_{x \to \infty}g(x) = \lim_{x \to \infty}\dfrac{1}{\dfrac{f(x)}{g(x)}}\cdot f(x) = \frac{1}{1}\cdot c = c$$ Note that the above manipulation rests on the fact that both $g(x), f(x)$ are non zero for large $x$. This is justified by the existence of limit $(1)$ which is non-zero.
On the other hand if we are given that $$\lim_{x \to \infty}g(x) = c\tag{3}$$ then $$\lim_{x \to \infty}f(x) = \lim_{x \to \infty}\frac{f(x)}{g(x)}\cdot g(x) = 1\cdot c = c$$ Note that the above manipulation rests on the fact that $g(x) \neq 0$ for all large values of $x$ and this is guaranteed by existence of the limit $(1)$.
