Question

Lim x→-1 x^m + 1/x^n + 1

Answer 1

I assume
`m,n` are integers to avoid (ir)rational powers of -1.

If
`m,n` are both even, or if
`m=n`, then

`\displaystyle \lim_(n\to-1) (x^m+1)/(x^n+1) = (1+1)/(1+1) = 1`

If
`m,n` are both odd and
`m\\eq n`, then we can factorize

`(x^m+1)/(x^n+1) = ((x+1)(x^(m-1) - x^(m-2) + \cdots - x + 1))/((x+1)(x^(n-1)-x^(n-2)+\cdots-x+1))`

Note that there are
`m` terms in the numerator and
`n` terms in the denominator.

In the limit, the factors of
`x+1` cancel and

`\displaystyle \lim_(x\to-1) (x^m+1)/(x^n+1) = \lim_(x\to-1) (x^(m-1) - x^(m-2) + \cdots - x + 1)/(x^(n-1)-x^(n-2)+\cdots-x+1) \\\\ ~~~~~~~~~~~~~~~~~~= (1-(-1)+1-(-1)+\cdots-(-1)+1)/(1-(-1)+1-(-1)+\cdots-(-1)+1) \\\\ ~~~~~~~~~~~~~~~~~~=(1+1+\cdots+1)/(1+1+\cdots+1) = \frac mn`

If
`m` is even and
`n` is odd, then we can only factorize the denominator and the discontinuity at
`x=-1` is nonremovable, so

`\displaystyle \lim_(x\to-1)(x^m+1)/(x^n+1) = \lim_(x\to-1) (x^m+1)/((x+1)(x^(n-1)-x^(n-2)+\cdots-x+1)) \\\\ ~~~~~~~~~~~~~~~~~~= \frac2m \lim_(x\to-1) \frac1{x+1}`

which does not exist.

If
`m` is odd and
`n` is even, then we can factorize the numerator so that

`\displaystyle \lim_(x\to-1)(x^m+1)/(x^n+1) = \lim_(x\to-1) ((x+1)(x^(m-1)-x^(m-2) +\cdots -x+1))/(x^n+1) \\\\ ~~~~~~~~~~~~~~~~~~= \frac{0m}2 = 0`