<s>[INST] How to solve $\lim _{x\to-\infty} \sqrt{x^2 + x - 1} + x$ I've tried everything with this question, once I multiply by the conjugate I can't simplify very well
$(x-1)/(\sqrt{x^2 + x - 1} - x)$
Then going to this line seems to mess with the limit, although it seems like I've done it right.
$(x-1)/(x(\sqrt{1 + 1/x - 1/x^2} - 1))$
I could also multiply by $(1/x)/(1/x)$ but again I get a similar problem
Once I cancel the $x$'s
$1/(\sqrt{1 + 1/x - 1/x^2} - 1) - 1/(\sqrt{x^2 + x - 1} - x)$
The first fraction becomes undefined
If someone could drop a stepping stone for me it would be very helpful
 [/INST] $$\begin{align*}
\lim_{x\to-\infty}\left(\sqrt{x^2+x-1}+x\right)&=\lim_{x\to-\infty}\left(\sqrt{x^2+x-1}+x\right)\frac{\sqrt{x^2+x-1}-x}{\sqrt{x^2+x-1}-x}\\[1ex]
&=\lim_{x\to-\infty}\frac{(x^2+x-1)-x^2}{\sqrt{x^2+x-1}-x}\\[1ex]
&=\lim_{x\to-\infty}\frac{x-1}{\sqrt{x^2}\sqrt{1+\frac{1}{x}-\frac{1}{x^2}}-x}\\[1ex]
&=\lim_{x\to-\infty}\frac{x-1}{|x|\sqrt{1+\frac{1}{x}-\frac{1}{x^2}}-x}\\[1ex]
&=\lim_{x\to-\infty}\frac{x-1}{-x\sqrt{1+\frac{1}{x}-\frac{1}{x^2}}-x}\\[1ex]
&=-\lim_{x\to-\infty}\frac{1-\frac{1}{x}}{\sqrt{1+\frac{1}{x}-\frac{1}{x^2}}+1}\\[1ex]
\end{align*}$$
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