How to prove $x^n$ has no limit for $x\leq -1$ Writing out the first few terms for $x=-1$:$$(-1)^1,(-1)^2,(-1)^3,...,(-1)^n$$ we can clearly see that the terms alternate between positive and negative and that since $-1\nleq x\ngeq1$it is clear that the sequence does not converge. My question is that how do I sum up my ideas into a rigorous proof?

For $(-1)^n$: Suppose it is converges to $l$ say. Then $$\vert (-1)^n -l \vert < \frac{1}{2}$$ for all $n \geq N$. 
For $n$ even, this says $$\vert 1 -l \vert < \frac{1}{2}$$ for all $n \geq N$
For $n$ odd, this says $$\vert -1 -l \vert < \frac{1}{2}$$ for all $n \geq N$
Now $$2=\vert 2 \vert =\vert 1+1\vert =\vert (1+l)+(1-l) \vert<1$$ which is absurd!
