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?

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.
