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?

$\sqrt{x^2}$ is a positive number and since we are taking the limit as $x\to -\infty$, we have $\sqrt{x^2} = -x$. Your limit will be negative since $x^2>x^2+x-1$ as $x\to-\infty$. 
