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

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.

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.