Counterexample regarding basic properties of limits Let $f, g$ be two functions defined on a deleted neighbourhood of
$a ∈ R$. If $g(x) \neq 0$ for every x on the deleted neighbourhood and $$\lim_{x\to a}{f(x)\over g(x)} = 1$$
then
$$\lim_{x\to a}{(f(x) - g(x))} = 0$$
I'm asked to prove or disprove this claim, along with others. My instinct in these type of questions is to start figuring out if there are any piecewise function that contradict the statements but I almost feel like I'm searching in the dark with no clear strategy.. is there an agreed upon process regarding searching for counter examples?
 A: Step One: Consider whether the problem can be simplified. In this case, we can eliminate the irrelevant distraction of $a$. Define $u=x-a$. Then instead of $\lim_{x \rightarrow a}f(x)$, we have $\lim_{u \rightarrow 0}f(u)$. Any proof or counterexample using the latter expression easily translates (no pun intended) to a solution of the former.
Step Two: State the negation. A statement of the form "if A then B" is false if there is a case where A is true but B is not. So we need a case where the first limit is 1, but the second is not zero.
Step Three: Enumerate the possibilities. There are four basic categories of limits: goes to zero, goes to a constant, goes to infinity, oscillates. The second two are both divergent on the real line, but the third is convergent on the extended real line. We're looking for a limit other than 0, so that leave a constant, infinity, or oscillates. The simplest is constant, and there isn't anything suggesting that one constant is better than another, so we can take it to be 1.
Step Four: Note the conditions/qualifications. The problem states that the functions need be defined only on a deleted neighborhood. Why would they include that detail? That suggests that we don't need to have them be defined for $u=0$. Why would they not be defined for $u=0$? Maybe they go to infinity there. What's the simplest function that goes to infinity? $\frac 1 u$. If one function is $\frac 1 u$, and the difference between that function and the other one is 1, then what is the other function? The simplest answer is $\frac 1 u+1$.
Step Five: Check whether you've accomplished the goal. It's generally easier to analyze a fraction when the more complicated expression is on top, so let's take $f(u) = \frac 1 u +1$ and $g(u) = \frac 1 u$. Now $\frac {f(u)}{g (u)} = u(\frac 1 u +1) = 1+u$. When we take the limit as $u$ goes to $0$, this indeed goes to $1$. And of course by construction the difference goes to $1$.
If you want to put this back in the original variable, you can substitute $u=x-a$ back in to get $f(x) =\frac 1 {x-a}+1, g(x)=\frac 1 {x-a}+1$, but I would expect that most instructors would find this last step trivial and the functions in term of $u$ to be sufficient.
If you combine Step One and Step Four, you might notice that if we take the substitution $u=\frac 1 {x-a}$, then the problem becomes to find two functions whose ration goes to $1$ when $u$ goes to infinity, but their difference doesn't go to $0$. If you're familiar with Big-O notation, that's asking for two functions in the same Big-O family whose difference doesn't go to zero, and $u+1$ and $u$ clearly satisfy that. When it comes to limits taking the independent variable to zero versus limits taking the independent variable to infinity, there are a lot of phenomena in one  that have an equivalent in the other, but one is easier to analyze than the other. If you're stuck with one of these types of limits, consider converting it to the other to see whether anything comes to mind.
A: What about
$$\begin{cases}
f(x) &= \frac{1}{(x-a)^2}+1\\
g(x) &= \frac{1}{(x-a)^2}
\end{cases}$$
The idea to find this counterexample is that if $\lim\limits_{x \to a} f(x)$ or $\lim\limits_{x \to a} g(x)$ exists and is finite, then the two limits are equal and therefore the limit of $f-g$ vanishes. So in the counterexample above, I looked at maps having non finite limits.
