Assuming $\lim\limits_{x \to a} \frac{f(x)}{g(x)}=1$ then I'm wonderin if the limit $\lim\limits_{x \to a} (f(x)- g(x))=0$, always?
You see, I was trying to see if the definitions for two asymptotic curves were equivalent. That is, the two definitions,
$\lim\limits_{x \to a} (f(x)- g(x))=0 \iff \lim\limits_{x \to a} \frac{f(x)}{g(x)}=1$ where $a \in [-\infty, \infty]$
After battling with this for the last half an hour, I have been able to produce a counterexample for $(\Rightarrow)$ which is pretty simple, take $f(x)=g(x)=0$. As for the $(\Leftarrow)$, I can also produce a counterexample for when $a ∈ \{±\infty\}$, take $f(x)$ and $g(x)$ to be polynomials of same degree like $f(x)=x^{99}, g(x)=x^{99}+7$, so the highest degree term dominates in the numerator and the denominator, hence limit of quotient is $1$ but difference is non-zero.
For the life of me, I can't come up with any pair of functions for the case $(\Leftarrow)$ when $a \in (-∞, ∞)$. I'm starting to believe it is actually true. Intuitively, it should be since it means near $a$, if $\frac{f(x)}{g(x)} ≈1$ then $f(x)≈ g(x)$ but my intuition has failed me enough times to know better than to rely on it. Any help?
Obviously, with this exercise, I have come to realize that the two definitions are not equivalent but I can't help but wonder why? Why are two definitions saying different things?
If $\lim_{x\to a} f(x) = L_1 < \infty$ and $\lim_{x\to a} g(x)= L_2$, it is easy to show that $\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L_1}{L_2}$. So, in this framework, if the limit of the quotient is 1, both limits are the same and the limit of the difference is zero.
Therefore, a counterexample must come from unbounded limits, as the ones proposed by @5xum.