Finding $\lim\limits_{n→∞}\frac{a^n+b^n}{a^n-b^n}$ with $a ≠ b$ 
Find $\lim\limits_{n→∞}\dfrac{a^n+b^n}{a^n-b^n}$ with $a ≠ b$.

Notice that I don't have $a>b$ or $b>a$, and also no $b≠0$ or $a≠0$, so I don't know if i even can use my solution.
My solution :
\begin{gather*}
\frac {a^n+b^n}{a^n-b^n} = \frac {a^n}{a^n-b^n} + \frac {b^n}{a^n-b^n}
= \frac{1}{1- \frac{b^n}{a^n}}-\frac {1}{\frac{a^n}{b^n}-1}\\
a>b \implies \lim\limits_{n→∞}\frac{b^n}{a^n}=0,\quad \lim\limits_{n→∞}\frac{a^n}{b^n}=\infty\\
\frac{1}{1- \frac{b^n}{a^n}}+\frac {1}{ \frac{a^n}{b^n}-1} = \frac {1}{1-0} + \frac{1}{\infty-1} = 1+\frac{1}{\infty}= 1+0=1\\
b>a \implies \lim\limits_{n→∞}\frac{b^n}{a^n}=\infty,\quad \lim\limits_{n→∞}\frac{a^n}{b^n}=0\\
\frac {1}{1- \frac{b^n}{a^n}}+\frac {1}{ \frac{a^n}{b^n}-1} = \frac {1}{1-\infty} + \frac {1}{0-1} = \frac{1}{-\infty}-1= 0-1=-1
\end{gather*}
Now I have two convergent solutions for different cases. What should I answer to the question mentioned above?
 A: Your answer should contain all possible cases, so in general, the answer will not be a single number.

Your solution, by the way, is correct, but it does not cover all cases. If $a>b$, then your proof that the limit is $1$ assumes that $a,b>0$! If $b<a<0$, then the limit $$\lim_{n\to\infty} \frac{a^n}{b^n}$$ is, in fact, $0$, not $\infty$.
Similarly, if $a<b$, then the limit is $-1$ according to your proof, but only if $a,b> 0$.
So, to conclude your examination of the limit, you must check what happens when values are negative.
A: Alternative approach:
First of all, I am interpreting the problem to represent that $n$ is restricted to the positive integers.  This is because, apparently, it is permissible to have either $a$ or $b$ negative.
If (for example), $a = -1.5$, you run into problems evaluating $a^n$ for $n \in \Bbb{R^+}$.  For example, is $(-1.5)^\pi$ a positive or negative number?
I avoid this by assuming that the problem intends that $n$ is restricted to $\Bbb{Z^+}.$

For fixed $a,b,~$ with $~n \in \Bbb{Z^+},~$ let $f(n) = \displaystyle \frac{a^n + b^n}{a^n - b^n}$.
$a = 0, b \neq 0 \implies f(n) = -1$, for all $n$, and for all $b$.
$b = 0 \implies f(n) = 1$, for all $n$, and for all $a$.
Therefore, without loss of generality, $a \neq 0 \neq b.$

$\underline{\textbf{Case 1:} ~|a| > |b|}$
Set $~r = \displaystyle \frac{b}{a} \implies -1 < r < 1 \implies \lim_{n \to \infty} r^n = 0$.
Then $f(n) = \displaystyle \frac{1 + r^n}{1 - r^n}$, which goes to $\displaystyle \frac{1 + 0}{1 - 0} = 1$.
Note that here, it is irrelevant whether $a$ is positive or negative, and irrelevant whether $b$ is positive or negative.

$\underline{\textbf{Case 2:} ~|a| < |b|}$
Set $~r = \displaystyle \frac{a}{b} \implies -1 < r < 1 \implies \lim_{n \to \infty} r^n = 0$.
Then $f(n) = \displaystyle \frac{r^n + 1}{r^n - 1}$, which goes to $\displaystyle \frac{0 + 1}{0 - 1} = -1$.

$\underline{\textbf{Case 3:} ~1 \leq |a| = |b|, ~a,b ~\text{are of opposite signs}}$
This is tricky.  When $n$ is odd, the numerator of $f(n)$ is $0$, while the denominator is increasing in absolute value.  So, the subsequence of just the terms corresponding to $n$ odd do go to $0$.
However, when $n$ is even, the numerator is increasing in absolute value, while the denominator is always $0$.  When the denominator of $f(n)$ is $0$, the fraction represented by $f(n)$ is undefined.
Here, it is unnecessary to examine the behavior of the subsequence represented by $n$ even.  The issue is the overall sequence, which is divergent, but does not go to infinity or go to minus infinity.

$\underline{\textbf{Case 4:} ~1 > |a| = |b|, ~a,b ~\text{are of opposite signs}}$
Superficially, this one may seem trickier than Case 3.
However, it is not trickier, because when $n$ is even, the denominator of $f(n)$ is flatly $0$, while the numerator is never $0$.  When $n$ is odd, the reverse is true (i.e. numerator only is $0$).
Therefore, the evaluation is identical to Case 3: the sequence is divergent, going to neither infinity or minus infinity.
