Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$ Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$.
My solution:
multiplying by: $\displaystyle\frac{\sqrt{x^2+x-1}-x}{\sqrt{x^2+x-1}-x}$ 
Which gives us: $\displaystyle\frac{x-1}{\sqrt{x^2+x-1}-x}$
dividing by $\sqrt{x^2}$ gives:
$\displaystyle \frac{1}{\sqrt{1}-1}$ 
which equals $1/0$
However, I double checked my answers, and this does not seem to be correct, am I making a mistake (perhaps when I take the $\sqrt{1}$ in the denominator of the last step?

Your mistake at this step:

dividing by $\sqrt{x^2}$ gives: $\displaystyle \frac{1}{\sqrt{1}-1}$ 

Since $x\to -\infty$, then $\sqrt{x^2}=-x$. So $\frac{x-1}{\sqrt{x^2+x-1}-x}=\frac{-1+\frac 1x}{\sqrt{1+\frac 1x-\frac{1}{x^2}}+1}.$
Take careful!
