<s>[INST] Calculate $\lim_{x \to \infty} x - \sqrt{x^2 + 2x}$ without derivations How can I calculate $\displaystyle \lim_{x \to \infty} x - \sqrt{x^2 + 2x}$?
Here is what I've done so far:
Multiplying by  $\displaystyle \frac{x + \sqrt{x^2 + 2x}}{x + \sqrt{x^2 + 2x}}$ 
I got  $\displaystyle \frac {-2x}{x+\sqrt{x^2 + 2x}}$
Multiplying this by $\displaystyle \frac{\frac{1}{x}}{\frac{1}{x}}$
I got $\displaystyle \frac{-2}{1+\frac{\sqrt{x^2 + 2x}}{x}}$
I know I'm very close to the answer wich is $-1$, but I have no idea what to do next. I can´t just say that $\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2 + 2x}}{x} = 1$, as far as I know...
 [/INST] Try:
$$
\frac{\sqrt{x^2 + 2x}}{x} = \sqrt{\frac{x^2 + 2x}{x^2}} = \sqrt{1 + \frac{2}{x}}
$$
By continuity of the square root function this means
$$
\lim_{x \to \infty} \frac{\sqrt{x^2 + 2x}}{x} = 1.
$$
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