I assume [tex]m,n[/tex] are integers to avoid (ir)rational powers of -1.
If [tex]m,n[/tex] are both even, or if [tex]m=n[/tex], then
[tex]\displaystyle \lim_{n\to-1} \frac{x^m+1}{x^n+1} = \frac{1+1}{1+1} = 1[/tex]
If [tex]m,n[/tex] are both odd and [tex]m\neq n[/tex], then we can factorize
[tex]\dfrac{x^m+1}{x^n+1} = \dfrac{(x+1)(x^{m-1} - x^{m-2} + \cdots - x + 1)}{(x+1)(x^{n-1}-x^{n-2}+\cdots-x+1)}[/tex]
Note that there are [tex]m[/tex] terms in the numerator and [tex]n[/tex] terms in the denominator.
In the limit, the factors of [tex]x+1[/tex] cancel and
[tex]\displaystyle \lim_{x\to-1} \frac{x^m+1}{x^n+1} = \lim_{x\to-1} \frac{x^{m-1} - x^{m-2} + \cdots - x + 1}{x^{n-1}-x^{n-2}+\cdots-x+1} \\\\ ~~~~~~~~~~~~~~~~~~= \dfrac{1-(-1)+1-(-1)+\cdots-(-1)+1}{1-(-1)+1-(-1)+\cdots-(-1)+1} \\\\ ~~~~~~~~~~~~~~~~~~=\frac{1+1+\cdots+1}{1+1+\cdots+1} = \dfrac mn[/tex]
If [tex]m[/tex] is even and [tex]n[/tex] is odd, then we can only factorize the denominator and the discontinuity at [tex]x=-1[/tex] is nonremovable, so
[tex]\displaystyle \lim_{x\to-1}\frac{x^m+1}{x^n+1} = \lim_{x\to-1} \frac{x^m+1}{(x+1)(x^{n-1}-x^{n-2}+\cdots-x+1)} \\\\ ~~~~~~~~~~~~~~~~~~= \frac2m \lim_{x\to-1} \frac1{x+1}[/tex]
which does not exist.
If [tex]m[/tex] is odd and [tex]n[/tex] is even, then we can factorize the numerator so that
[tex]\displaystyle \lim_{x\to-1}\frac{x^m+1}{x^n+1} = \lim_{x\to-1} \frac{(x+1)(x^{m-1}-x^{m-2} +\cdots -x+1)}{x^n+1} \\\\ ~~~~~~~~~~~~~~~~~~= \frac{0m}2 = 0[/tex]
