<s>[INST] Limit of: $ -x+\sqrt{x^2+x} $ for $ x\to\infty $ Why the solution of this limit: $$\lim_{x\to\infty}-x+\sqrt{x^2+x} $$ is $\frac{1}{2}$?
I tried to do De L'hopital to $$\frac{-x^2+x\sqrt{x^2+x}}{x}$$ but whitout success.
 [/INST] $$\begin{align}
-x+\sqrt{x^2+x}&=-x+\sqrt{x^2+x}\left(\frac{x+\sqrt{x^2+x}}{x+\sqrt{x^2+x}}\right)\\\\
&=\frac{x}{x+\sqrt{x^2+x}}\\\\
&=\frac{1}{1+\sqrt{1+1/x}}\to 1/2\,\,\text{as}\,\,x\to \infty
\end{align}$$
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