Showing that $\lim_{x \to -\infty}(x+{1\over x}) = -\infty$ My attempt. I' not sure about this, but I think the definition for this limit will be: 
$\forall M <0 \space \exists N < 0, \space x < N \Rightarrow f(x) < M$
We start off knowing that $x < N$ $\Rightarrow$ $x^2 > N^2\Rightarrow {1\over x^2 } + 1 < {1 \over N^2} + 1 \Rightarrow x({1\over x^2 } + 1) < N({1 \over N^2} + 1)$. Now, since ${1\over N^2} + 1 > 1$ and $N < 0$, we have that $N({1\over N^2} + 1 ) < N$. Thefore, $x({1\over x^2} + 1) < N$, which mean that we will choose N = M in our proof.
Proof:
Pick N = M. Since $x < N \Rightarrow x({1\over x^2 } + 1) < N({1 \over N^2} + 1)< N = M$, as required. 
Is this correct?

You wanted to evaluate the limit from first principles, so
Yes, you want to prove :
$\forall M <0,\space \exists N < 0, \space x < N \Rightarrow f(x) < M$
"We start off knowing that $x < N$ $\Rightarrow$ $x^2 > N^2\Rightarrow {1\over x^2 } + 1 < {1 \over N^2} + 1$" This is correct.
$\Rightarrow x({1\over x^2 } + 1) < N({1 \over N^2} + 1)$.
This is wrong! Consider $x=-0.5,N=-0.4,x<N$, but 
$\Rightarrow x({1\over x^2 } + 1)=-2.5 > N({1 \over N^2} + 1)=-2.9!$
You are making an invalid comparison here.
Instead you can simply do this:
Proof:
Pick N = M. Since $x < N,x^2>xM$,
$x^2+1>x^2>xM$, since obviously $x\neq 0$,
$f(x)=x+\frac{1}{x}<x<M$, as required.
